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Some Special Integrals

Class 12th Mathematics RS Aggarwal Solution
Exercise 14a
  1. integrate {dx}/{ (1-9x)^{2} } Evaluate:
  2. integrate {dx}/{ ( 25-4x^{2} ) } Evaluate:
  3. integrate {dx}/{ ( x^{2} + 16 ) } Evaluate:
  4. integrate {dx}/{ ( 4+9x^{2} ) } Evaluate:
  5. integrate {dx}/{ ( 50+2x^{2} ) } Evaluate:
  6. integrate {dx}/{ ( 16x^{2} - 25 ) } Evaluate:
  7. integrate { ( x^{2} - 1 ) }/{ ( x^{2} + 4 ) } dx Evaluate:
  8. integrate { x^{2} }/{ ( 9+4x^{2} ) } dx Evaluate:
  9. integrate { e^{x} }/{ ( e^{2x} + 1 ) } dx Evaluate:
  10. integrate {sinx}/{ (1+cos^{2}x) } dx Evaluate:
  11. integrate {cosx}/{ (1+sin^{2}x) } dx Evaluate:
  12. integrate { 3x^{5} }/{ ( 1+x^{12} ) } dx Evaluate:
  13. integrate { 2x^{3} }/{ ( 4+x^{8} ) } dx Evaluate:
  14. integrate {dx}/{ ( e^{x} + e^{-x} ) } Evaluate:
  15. integrate {x}/{ ( 1-x^{4} ) } dx Evaluate:
  16. integrate { x^{2} }/{ ( a^{6} - x^{6} ) } dx Evaluate:
  17. integrate {dx}/{ ( x^{2} + 4x+8 ) } Evaluate:
  18. integrate {dx}/{ ( 4x^{2} - 4x+3 ) } Evaluate:
  19. integrate {dx}/{ ( 2x^{2} + x+3 ) } Evaluate:
  20. integrate {dx}/{ ( 2x^{2} - x-1 ) } Evaluate:
  21. integrate {dx}/{ ( 3-2x-x^{2} ) } Evaluate:
  22. integrate {x}/{ ( x^{2} + 3x+2 ) } dx Evaluate:
  23. integrate { (x-3) }/{ ( x^{2} + 2x-4 ) } dx Evaluate:
  24. integrate { (2x-3) }/{ ( x^{2} + 3x-18 ) } dx Evaluate:
  25. integrate { x^{2} }/{ ( x^{2} + 6x-3 ) } dx Evaluate:
  26. integrate { (2x-1) }/{ ( 2x^{2} + 2x+1 ) } dx Evaluate:
  27. integrate { (1-3x) }/{ ( 3x^{2} + 4x+2 ) } dx Evaluate:
  28. integrate {2x}/{ ( 2+x-x^{2} ) } dx Evaluate:
  29. integrate {dx}/{ (1+cos^{2}x) } Evaluate:
  30. integrate {dx}/{ (2+sin^{2}x) } Evaluate:
  31. integrate {dx}/{ (a^{2}cos^{2}x+b^{2}sin^{2}x) } Evaluate:
  32. integrate {dx}/{ (cos^{2}x-3sin^{2}x) } Evaluate:
  33. integrate {dx}/{ (sin^{2}x-4cos^{2}x) } Evaluate:
  34. integrate {dx}/{ (sinxcosx+2cos^{2}x) } Evaluate:
  35. integrate {sin2x}/{ ( sin^{4}x+cos^{4}x } dx Evaluate:
  36. integrate { (2sin2phi -cosphi) }/{ (6-cos^{2}phi-4sinphi) } d phi Evaluate:…
  37. integrate {dx}/{ (sinx-2cosx) (2sinx+cosx) } Evaluate:
  38. integrate { ( 1-x^{2} ) }/{ ( 1+x^{4} ) } dx Evaluate:
  39. integrate { ( x^{2} + 1 ) }/{ ( x^{4} + x^{2} + 1 ) } dx Evaluate:…
  40. integrate {dx}/{ (sin^{4}x+cos^{4}x) } Evaluate:
Exercise 14b
  1. integrate {dx}/{ root { 16-x^{2} } } Evaluate:
  2. integrate {dx}/{ root { 1-9x^{2} } } Evaluate:
  3. integrate {dx}/{ root { 15-8x^{2} } } Evaluate:
  4. integrate {dx}/{ root { x^{2} - 4 } } Evaluate:
  5. integrate {dx}/{ root { 4x^{2} - 1 } } Evaluate:
  6. integrate {dx}/{ root { 9x^{2} - 7 } } Evaluate:
  7. integrate {dx}/{ root { x^{2} - 9 } } Evaluate:
  8. integrate {dx}/{ root { 1+4x^{2} } } Evaluate:
  9. integrate {dx}/{ root { 9+4x^{2} } } Evaluate:
  10. integrate {x}/{ root { 9-x^{4} } } dx Evaluate:
  11. integrate { 3x^{2} }/{ root { 9-16x^{6} } } dx Evaluate:
  12. integrate {sec^{2}x}/{ root {16+tan^{2}x} } dx Evaluate:
  13. integrate {sinx}/{ root {4+cos^{2}x} } dx Evaluate:
  14. integrate {cosx}/{ root {9sin^{2}x}-1 } dx Evaluate:
  15. integrate { e^{x} }/{ root { 4+e^{2x} } } dx Evaluate:
  16. integrate { 2e^{x} }/{ root { 4-e^{2x} } } dx Evaluate:
  17. integrate {dx}/{ root { 1-e^{x} } } Evaluate:
  18. integrate root { {a-x}/{a+x} } dx Evaluate:
  19. integrate {dx}/{ root { x^{2} + 6x+5 } } Evaluate:
  20. integrate {dx}/{ root { (2-x)^{2} + 1 } } Evaluate:
  21. integrate {dx}/{ root { (x-3)^{2} + 1 } } Evaluate:
  22. integrate {dx}/{ root { x^{2} - 6x+10 } } Evaluate:
  23. integrate {dx}/{ root { 2+2x-x^{2} } } Evaluate:
  24. integrate {dx}/{ root { 8-4x-2x^{2} } } Evaluate:
  25. integrate {dx}/{ root { 16-6x-x^{2} } } Evaluate:
  26. integrate {dx}/{ root { 7-6x-x^{2} } } Evaluate:
  27. integrate {dx}/{ root { x-x^{2} } } Evaluate:
  28. integrate {dx}/{ root { 8+2x-x^{2} } } Evaluate:
  29. integrate {dx}/{ root { x^{2} - 3x+2 } } Evaluate:
  30. integrate {dx}/{ root { 2x^{2} + 3x-2 } } Evaluate:
  31. integrate {dx}/{ root { 2x^{2} + 4x+6 } } Evaluate:
  32. integrate {dx}/{ root { 1+2x-3x^{2} } } Evaluate:
  33. integrate {dx}/{ root {x} sqrt{5-x} } Evaluate:
  34. integrate {dx}/{ root { 3+4x-2x^{2} } } Evaluate:
  35. integrate { x^{2} }/{ root { x^{6} + 2x^{3} + 3 } } dx Evaluate:…
  36. integrate { (2x+3) }/{ root { x^{2} + x+1 } } dx Evaluate:
  37. integrate { (5x+3) }/{ root { x^{2} + 4x+10 } } dx Evaluate:
  38. integrate { (4x+3) }/{ root { 2x^{2} + 2x-3 } } Evaluate:
  39. integrate { (3-2x) }/{ root { 2+x-x^{2} } } dx Evaluate:
  40. integrate { (x+2) }/{ root { 2x^{2} + 2x-3 } } dx Evaluate:
  41. integrate { (3x+1) }/{ root { 5-2x-x^{2} } } dx Evaluate:
  42. integrate { (6x+5) }/{ root { 6+x-2x^{2} } } dx Evaluate:
  43. integrate root { {1+x}/{x} } dx Evaluate:
  44. integrate { (x+2) }/{ root { x^{2} + 5x+6 } } dx Evaluate:
Exercise 14c
  1. integrate root { 4-x^{2} } dx Evaluate the following integrals:…
  2. integrate root { 4-9x^{2} } dx Evaluate the following integrals:…
  3. integrate root { x^{2} - 2 } dx Evaluate the following integrals:…
  4. integrate root { 2x^{2} - 3 } dx Evaluate the following integrals:…
  5. integrate root { x^{2} + 5 } dx Evaluate the following integrals:…
  6. integrate root { 4x^{2} + 9 } dx Evaluate the following integrals:…
  7. integrate root { 3x^{2} + 4 } dx Evaluate the following integrals:…
  8. integrate cosxroot {9-sin^{2}x}dx Evaluate the following integrals:…
  9. integrate root { x^{2} - 4x+2 } dx Evaluate the following integrals:…
  10. integrate root { x^{2} + 6x-4 } dx Evaluate the following integrals:…
  11. integrate root { 2x-x^{2} } dx Evaluate the following integrals:…
  12. integrate root { 1-4x-x^{2} } dx Evaluate the following integrals:…
  13. integrate root { 2ax-x^{2} } dx Evaluate the following integrals:…
  14. integrate root { 2x^{2} + 3x+4 } dx Evaluate the following integrals:…
  15. integrate root { x^{2} + x } dx Evaluate the following integrals:…
  16. integrate root { x^{2} + x+1 } dx Evaluate the following integrals:…
  17. integrate (2x-5) root { x^{2} - 4x+3 } dx Evaluate the following integrals:…
  18. integrate (x+2) root { x^{2} + x+1 } dx Evaluate the following integrals:…
  19. integrate (x-5) root { x^{2} + x } dx Evaluate the following integrals:…
  20. integrate (4x+1) root { x^{2} - x-2 } dx Evaluate the following integrals:…
  21. integrate (x+1) root { 2x^{2} + 3 } dx Evaluate the following integrals:…
  22. integrate x root { 1+x-x^{2} } dx Evaluate the following integrals:…
  23. Evaluate the following integrals:
  24. integrate (6x+5) root { 6+x-2x^{2} } dx Evaluate the following integrals:…
  25. integrate (x+1) root { 1-x-x^{2} } dx Evaluate the following integrals:…
  26. integrate (x-3) root { x^{2} + 3x-18 } dx Evaluate the following integrals:…
Exercise 14a
Question 1.

Evaluate:
Answer:

To find:


Formula Used:


Let y = (1 – 9x) … (1)


Differentiating with respect to x,



i.e., dy = -9 dx


Substituting in the equation to evaluate,






Simplifying and substituting the value of y from (1),




Therefore,
Question 2.

Evaluate:
Answer:

To find:


Formula Used:


Given equation =


… (1)


Here


Therefore, (1) becomes




Therefore,
Question 3.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,



Here a = 4



Therefore,
Question 4.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,




Here




Therefore,
Question 5.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,




Here a = 5



Therefore,
Question 6.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,




Here




Therefore,
Question 7.

Evaluate:
Answer:

To find:


Formula Used:


Given equation can be rewritten as the following:





Here a = 2,



Therefore,
Question 8.

Evaluate:
Answer:

To find:


Formula Used:


Given equation can be rewritten as the following:





Here ,




Therefore,
Question 9.

Evaluate:
Answer:

To find:


Formula Used:


Let y = ex … (1)


Differentiating both sides, we get


dy = ex dx


Substituting in given equation,



⇒ tan-1 y


From (1),


⇒ tan-1 (ex)


Therefore,
Question 10.

Evaluate:
Answer:

To find:


Formula Used:


Let y = cos x … (1)


Differentiating both sides, we get


dy = –sin x dx


Substituting in given equation,



⇒ – tan-1 y


From (1),


⇒ –tan-1 (cos x)


Therefore,
Question 11.

Evaluate:
Answer:

To find:


Formula Used:


Let y = sin x … (1)


Differentiating both sides, we get


dy = cos x dx


Substituting in given equation,



⇒ tan-1 y


From (1),


⇒ tan-1 (sin x)


Therefore,
Question 12.

Evaluate:
Answer:

To find:


Formula Used:


Let y = x6 … (1)


Differentiating both sides, we get


dy = 6x5 dx


Substituting in given equation,




From (1),



Therefore,
Question 13.

Evaluate:
Answer:

To find:


Formula Used:


Let y = x4 … (1)


Differentiating both sides, we get


dy = 4x3 dx


Substituting in given equation,





From (1),



Therefore,
Question 14.

Evaluate:
Answer:

To find:


Formula Used:


Given equation is:


… (1)


Let y = ex … (1)


Differentiating both sides, we get


dy = ex dx


Substituting in (1),



⇒ tan-1 y


From (1),


⇒ tan-1 (ex)


Therefore,
Question 15.

Evaluate:
Answer:

To find:


Formula Used:


Let y = x2 … (1)


Differentiating both sides, we get


dy = 2x dx


Substituting in given equation,



Here a = 1,




From (1),



Therefore,
Question 16.

Evaluate:
Answer:

To find:


Formula Used:


Let y = x3 … (1)


Differentiating both sides, we get


dy = 3x2 dx


Substituting in given equation,






From (1),



Therefore,
Question 17.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,



… (1)


Let y = x + 2 … (2)


Differentiating both sides,


dy = dx


Substituting in (1),



Here a = 2,



From (2),



Therefore,
Question 18.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,


… (1)


Let y = 2x – 1 … (2)


Differentiating both sides,


dy = 2dx


Substituting in (1),



Here a = √2,



From (2),



Therefore,
Question 19.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,



… (1)


Let … (2)


Differentiating both sides,


dy = √2 dx


Substituting in (1),



Here



From (2),



Therefore,
Question 20.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,





… (1)


Let … (2)


Differentiating both sides,


dy = √2 dx


Substituting in (1),



Here




From (2),





Therefore,
Question 21.

Evaluate:
Answer:

To find:


Formula Used:


Rewriting the given equation,




… (1)


Let y = x + 1 … (2)


Differentiating both sides wrt x,


dy = dx


Substituting in (1),




Here a = 2,



From (2),



Therefore,
Question 22.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Using partial fractions,



x = A (2x + 3) + B


Equating the coefficients of x,


1 = 2A



Also, 0 = 3A + B



Therefore, the given equation becomes,







Therefore,
Question 23.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Using partial fractions,



x – 3 = A (2x + 2) + B


Equating the coefficients of x,


1 = 2A



Also, -3 = 2A + B


⇒ B = -4


Substituting in the given equation,






Therefore,
Question 24.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Using partial fractions,



2x – 3 = A(2x + 3) + B


Equating the coefficients of x,


2 = 2A


A = 1


Also, -3 = 3A + B


⇒ B = -6


Substituting in the given equation,



… (1)


Let I



Here a



… (2)


Substituting (2) in (1),



Therefore,
Question 25.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Given equation can be rewritten as following:





Let I … (2)


Using partial fractions,



6x – 3 = A (2x + 6) + B


Equating the coefficients of x,


6 = 2A


A = 3


Also, -3 = 6A + B


⇒ B = -21


Substituting in (1),





I


Therefore,
Question 26.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Using partial fractions,



2x – 1 = A (4x + 2) + B


Equating the coefficients of x,


2 = 4A


A


Also, -1 = 2A + B


⇒ B = -2


Substituting in the given equation,




Let I … (1)






Here



⇒ 2tan-1(2x + 1) + C


Substituting in (1) and combining with original equation,



Therefore,
Question 27.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Rewriting the given equation,



Using partial fractions,



3x – 1 = A (6x + 4) + B


Equating the coefficients of x,


3 = 6A


A


Also, -1 = 4A + B


⇒ B = -3


Substituting in the original equation,




Let I





Here




Substituting in (1) and combining with original equation,



Therefore,
Question 28.

Evaluate:
Answer:

To find:


Formula Used:


1.


2.


Rewriting the given equation,



Using partial fractions,



x = A (2x – 1) + B


Equating the coefficients of x,


1 = 2A


A


Also, 0 = -A + B


B


Substituting in the original equation,




Let I




Here




Substituting for I and combining with the original equation,



Therefore,



or
Question 29.

Evaluate:
Answer:

To find:


Formula Used:


1.


2. sec2 x = 1 + tan2 x


Dividing the given equation by cos2x in the numerator and denominator gives us,


… (1)


Let y = tan x


dy = sec2 x dx … (2)


Also, y2 = tan2 x


i.e., y2 = sec2 x – 1


sec2 x = y2 + 1 … (3)


Substituting (2) and (3) in (1),





Since y = tan x,



Therefore,
Question 30.

Evaluate:
Answer:

To find:


Formula Used:


1.


2. sec2 x = 1 + tan2 x


Dividing the given equation by cos2x in the numerator and denominator gives us,


… (1)


Let y = tan x


dy = sec2 x dx … (2)


Also, y2 = tan2 x


i.e., y2 = sec2 x – 1


sec2 x = y2 + 1 … (3)


Substituting (2) and (3) in (1),






Since y = tan x,



Therefore,
Question 31.

Evaluate:
Answer:

To find:


Formula Used:


1. Sec2 x = 1 + tan2 x


2.


Dividing by cos2 x in the numerator and denominator,



Let y = tan x


dy = sec2 x dx


Therefore,





Since y = tan x,



Therefore,
Question 32.

Evaluate:
Answer:

To find:


Formula Used:


1. sec2 x = 1 + tan2 x


2.


Dividing by cos2 x in the numerator and denominator,



Let y = tan x


dy = sec2 x dx


Therefore,






Since y = tan x,



Therefore,
Question 33.

Evaluate:
Answer:

To find:


Formula Used:


1. sec2 x = 1 + tan2 x


2.


Dividing by cos2 x in the numerator and denominator,



Let y = tan x


dy = sec2 x dx


Therefore,




Since y = tan x,



Therefore,
Question 34.

Evaluate:
Answer:

To find:


Formula Used:


1. sec2 x = 1 + tan2 x


2.


Dividing by cos2 x in the numerator and denominator,



Let y = tan x


dy = sec2 x dx


Therefore,



⇒ log |y + 2| + C


Since y = tan x,


⇒ log |tan x + 2| + C


Therefore,
Question 35.

Evaluate:
Answer:

To find:


Formula Used:


1. sec2 x = 1 + tan2 x


2.


3. sin 2x = 2 sin x cos x


Rewriting the given equation,



Dividing by cos4 x in the numerator and denominator,



Let y = tan x


dy = sec2 x dx


Therefore,



Let z = y2


dz = 2y dy



⇒ tan-1 z + C


Since z = y2,


⇒ tan-1(y2) + C


Since y = tan x,


⇒ tan-1(tan2 x) + C


Therefore,
Question 36.

Evaluate:
Answer:

To find:


Formula Used:


1. sec2 x = 1 + tan2 x


2.


3. sin 2x = 2 sin x cos x


Rewriting the given equation,





Let y = sin ϕ


dy = cos ϕ dϕ


Substituting in the original equation,


… (1)


Using partial fraction,



4y – 1 = A (2y - 4) + B


Equating the coefficients of y,


4 = 2A


A = 2


Also, -1 = -4A + B


B = 7


Substituting in (1),




⇒ 2 log |y2 – 4y + 5| + 7 tan-1(y – 2) + C


But y = sin ϕ


⇒ 2 log |sin2ϕ – 4 sin ϕ + 5| + 7 tan-1(sin ϕ – 2) + C


Therefore,
Question 37.

Evaluate:
Answer:

To find:


Formula Used:


1. sec2 x = 1 + tan2 x


2.


Dividing by cos2 x in the numerator and denominator,



Let y = tan x


dy = sec2 x dx


Therefore,


… (1)


Let



1 = A (2y + 1) + B(y – 2)


When y = 0,


1 = A – 2B … (2)


When y = 1,


1 = 3A – B ⇒ 2 = 6A – 2B … (3)


Solving (2) and (3),


1 = 5A


A


So, B


(1) becomes,




Since y = tan x,




Therefore,
Question 38.

Evaluate:
Answer:

To find:


Formula used:


On dividing by x2 in the numerator and denominator of the given equation,





Let


Differentiating wrt x,



Substituting in the original equation,




Substituting for and taking reciprocal of the value within logarithm, we get




Therefore,
Question 39.

Evaluate:
Answer:

To find:


Formula used:


On dividing by x2 in the numerator and denominator of the given equation,




Let


Differentiating wrt x,



Substituting in the original equation,




Substituting for




Therefore,
Question 40.

Evaluate:
Answer:

To find:


Formula used:


1. sec2 x = 1 + tan2x


2.


Dividing by cos4 x in the numerator and denominator of the given equation,




Let y = tan x


dy = sec2 x dx


Substituting in the original equation,



Dividing by y2 in the numerator and denominator,





Let



Therefore,




Substituting for z,




Substituting for y = tan x,



Therefore,
Exercise 14b
Question 1.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 2.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant






, c being the integrating constant
Question 3.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant






, c being the integrating constant
Question 4.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 5.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 6.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 7.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 8.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 9.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 10.

Evaluate:
Answer:

Tip – d(x2) = 2xdx i.e. xdx = (1/2)×d(x2)


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 11.

Evaluate:
Answer:

Tip – d(x3) = 3x2dx so, d(4x3) = 4×3x2dx i.e 3x2dx = (1/4)d(2x3)


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 12.

Evaluate:
Answer:

Tip – d(tanx) = sec2xdx


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 13.

Evaluate:
Answer:

Tip – d(cosx) = - sinxdx i.e. sinxdx = - d(cosx)


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 14.

Evaluate:
Answer:

Tip – d(sinx) = cosxdx so, d(3sinx) = 3cosxdx i.e. cosxdx = (1/3)d(3sinx)


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 15.

Evaluate:
Answer:

Tip – d(ex) = exdx


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 16.

Evaluate:
Answer:

Tip – d(ex) = exdx


Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 17.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant






Tip – Assuming e - (x/2) = a, - (1/2) e - (x/2)dx = da i.e. e - (x/2)dx = - 2da





, c being the integrating constant
Question 18.

Evaluate:
Answer:

Tip – Taking ,


and


i.e






Formula to be used -












, c being the integrating constant
Question 19.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 20.

Evaluate:
Answer:

Tip – d(2 - x) = - dx i.e. dx = - d(2 - x)


Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 21.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant




, c being the integrating constant
Question 22.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 23.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 24.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant






, c being the integrating constant
Question 25.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 26.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 27.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant






, c being the integrating constant
Question 28.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 29.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 30.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 31.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 32.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant







, c being the integrating constant
Question 33.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant






, c being the integrating constant
Question 34.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant







, c being the integrating constant
Question 35.

Evaluate:
Answer:

Tip – d(x3) = 3x2dx i.e. x2dx = (1/3)d(x3)


Formula to be used - where c is the integrating constant





, c being the integrating constant
Question 36.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming x2 + x + 1 = a2, (2x + 1)dx = 2ada











, c is the integrating constant
Question 37.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming x2 + 4x + 10 = a2, (2x + 4)dx = 2ada











, c is the integrating constant
Question 38.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming 2x2 + 2x - 3 = a2, (4x + 2)dx = 2ada












, c is the integrating constant
Question 39.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming 2 + x - x2 = a2, (1 - 2x)dx = 2ada











, c is the integrating constant
Question 40.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming 2x2 + 2x - 3 = a2, (4x + 2)dx = 2ada












, c is the integrating constant
Question 41.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming 5 - 2x - x2 = a2, ( - 2 - 2x)dx = 2ada i.e. (x + 1)dx = - ada










, c is the integrating constant
Question 42.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Assuming 6 + x - 2x2 = a2, (1 - 4x)dx = 2ada i.e. (4x - 1)dx = - 2ada












, c is the integrating constant
Question 43.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant







Tip – Taking x2 + x = a2 , (2x + 1)dx = 2ada










, c is the integrating constant
Question 44.

Evaluate:
Answer:

Formula to be used - where c is the integrating constant





Tip – Taking x2 + 5x + 6 = a2 , (2x + 5)dx = 2ada










, c is the integrating constant
Exercise 14c
Question 1.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = + + C


Since is of the form ,


Hence, = + + C


= + + C


= + 2 + C


Therefore, = + 2 + C
Question 2.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = + + C


Since is of the form ,


Hence, = + + C


= + + C


= + + C


Therefore, = + + C
Question 3.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = log |x +|+ C


Since is of the form ,


Hence, = log |x +|+ C


= log |x +|+ C


= log |x +|+ C


Therefore, = log |x +|+ C
Question 4.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = log |x +|+ C


Since is of the form ,


Hence, = log |x +|+C


= log |x +|+ C


= log |x +|+ C


Therefore, = log |x +|+ C
Question 5.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = log |x +|+ C


Since is of the form ,


Hence, = log |x +|+ C


= log |x +|+ C


Therefore, = log |x +|+ C
Question 6.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = log |x +|+ C


Since is of the form ,


Hence, = log |2x +|+ C


= log |2x + | + C


= log |2x +|+ C


Therefore, = log |2x +|+ C
Question 7.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


Formula Used: = log |x +|+ C


Since is of the form ,


Hence, = log |x +|+ C


= log |x + | + C


= log |x +|+ C


Therefore, = log |x +|+ C
Question 8.

Evaluate the following integrals:
Answer:

To Find :


Now, let sin x = t


cosx dx = dt


Therefore, can be written as


Formula Used: = + + C


Since , is in the form of with t as a variable instead of x .


= + + C


= + + C


Now since sin x = t and cosx dx = dt


= + + C
Question 9.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


i.e.,


Here , let x – 2 = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


= 2 log |y +|+ C


Since , x – 2 = y and dx = dy


= 2 log |(x-2) +|+C Therefore,


= 2 log |(x - 2) +|+ C
Question 10.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


i.e,


Here , let x + 3 = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x + 3 = y and dx = dy


= log |(x + 3) +|+C


Therefore,


= log |(x + 3) +|+C
Question 11.

Evaluate the following integrals:
Answer:

To Find : dx


Now, can be written as


i.e,


Let x – 1 = y dx = dy


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable,


Hence = + + C


= + + C


Here we have x – 1 = y and dx = dy


= + + C


Therefore , = + + C
Question 12.

Evaluate the following integrals:
Answer:

To Find : dx


Now, can be written as


i.e,


Let x + 2 = y dx = dy


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable,


Hence = + + C


= + + C


Here we have x + 2 = y and dx = dy


= + + C


Therefore , = + + C
Question 13.

Evaluate the following integrals:
Answer:

To Find : dx


Now, can be written as


i.e,


Let x - a = y dx = dy


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable,


Hence = + + C


= + + C


Here we have x - a = y and dx = dy


= + + C


Therefore , = + + C
Question 14.

Evaluate the following integrals:
Answer:

To Find :


Now , consider =


=


=


=


Let x + = y dx = dy


Hence becomes


Formula Used: = log |x +|+ C


Now consider which is in the form of with change in variable.


= log |y +|+ C


= log |y +|+ C


Since x + = y and dx = dy


= log | +|+ C


Now , = log | +|+ C


Therefore,


= log | +|+ C
Question 15.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


i.e,


Here , let x + = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x + = y and dx = dy


= log |(x + ) +|+C


Therefore,


= log |x + +|+C
Question 16.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


i.e,


Here , let x + = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x + = y and dx = dy


= log |(x + ) +|+C


Therefore,


= log |x + +|+C
Question 17.

Evaluate the following integrals:
Answer:

To Find :


Now, let be written as (2x – 4) -1 and split


Therefore ,


=


=


Now solving,


Let = u dx =


Thus, becomes


Now , = = =


=


Now solving,


=


=


Let x – 2 = y dx = dy


Then becomes


Formula Used: = log |x +|+ C


Since is in the form of with change in variable.


Hence = log |y +|+ C


= log |y +|+ C


Now, since x – 2 = y and dx = dy


= log |(x-2) +|+ C


Hence = log |(x-2) +|+ C


Therefore , =


log |(x-2) +|+ C


=


log |x - 2 +|+ C
Question 18.

Evaluate the following integrals:
Answer:

To Find :


Now, let x + 2 be written as (2x + 1) + and split


Therefore ,


=


=


Now solving,


Let = u dx =


Thus, becomes


Now , = = ) =


=


Now solving ,


Now, can be written as


i.e,


Here , let x + = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x + = y and dx = dy


= log |(x + ) +|+C


Therefore,


= log |x + +|+C


Hence ,


= + log |(x + )+|+C


Therefore , = + log |(x + )+|+C
Question 19.

Evaluate the following integrals:
Answer:

To Find :


Now, let x - 5 be written as (2x + 1) - and split


Therefore ,


=


=


Now solving,


Let = u dx =


Thus, becomes


Now , = = ) =


=


Now solving,


Now, can be written as


i.e,


Here , let x + = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x + = y and dx = dy


= log |(x + ) +|+C


Therefore,


= log |x + +|+C


Now ,


= - log |x + +|+C


Therefore ,


= - log |x + +|+C
Question 20.

Evaluate the following integrals:
Answer:

To Find :


Now, let 4x + 1 be written as 2(2x - 1) + 3 and split


Therefore ,


=


=


Now solving,


Let = u dx =


Thus, becomes


Now , = = ) =


=


Now solving,


Now, can be written as


i.e,


Here , let x - = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x - = y and dx = dy


= log |(x - ) +|+C


Therefore,


= log |x - +|+C


Hence ,


= + log |x - +|+C


Therefore ,


= + log |x - +|+C
Question 21.

Evaluate the following integrals:
Answer:

To Find :


Now, can be written as


=


=


Now solving,


Let = u dx =


Thus, becomes


Now , = = ) =


=


Now solving,


Now, can be written as


Formula Used: = log |x +|+ C


Since is of the form .


= log |x +|+ C


= log |x +|+ C


Therefore,


= + log |x +|+ C


Hence ,


= + log |x +|+ C
Question 22.

Evaluate the following integrals:
Answer:

To Find :


Now, let x be written as - (1 - 2x) and split


Therefore ,


=


= +


Now solving,


Let = u dx =


Thus,becomes


Now , = = ) =


=


Now solving,


can be written as


i.e, =


let 2x – 1 = y dx =


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable .


Hence, = + + C


= + + C


Since , 2x – 1 = y and dx =


Therefore,


= + + C


i.e, = + + C


hence , = + = = + + + C
Question 23.

Evaluate the following integrals:
Answer:

To Find :


Now, let 2x - 5 be written as (2x – 3) -2 and split


Therefore ,


=


= -


Now solving,


Let = u dx =


Thus,becomes


Now , = = ) =


=


Now solving,


can be written as


i.e,


let x – = y dx = dy


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable .


Hence, = + + C


= + + C


Since , x – = y and dx = dy


Therefore,


= + + C


i.e, = + + C


hence ,


= - = - - + C
Question 24.

Evaluate the following integrals:
Answer:

To Find :


Now, let 6x + 5 be written as - (1 - 4x)and split


Therefore ,


=


= +


Now solving,


Let = u dx =


Thus,becomes


Now , = = ) =


=


Now solving,


can be written as


i.e,


let = y dx =


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable .


Hence, = + + C


= + + C


Since , = y and dx =


Therefore,


= + + C


i.e, = + + C


hence ,


= + = + + + C
Question 25.

Evaluate the following integrals:
Answer:

To Find :


Now, let x + 1 be written as - (-2x - 1) and split


Therefore ,


=


= +


Now solving,


Let = u dx =


Thus,becomes


Now , = = ) =


=


Now solving,


can be written as


i.e,


let = y dx = dy


Therefore , becomes


Formula Used: = + + C


Since is of the form with change in variable .


Hence, = + + C


= + + C


Since , = y and dx = dy


Therefore,


= + + C


i.e, = + + C


hence ,


= + = + + + C
Question 26.

Evaluate the following integrals:
Answer:

To Find :


Now, let x - 3 be written as (2x + 3) - and split


Therefore ,


=


= -


Now solving,


Let = u dx =


Thus,becomes


Now , = = ) =


=


Now solving,


can be written as


i.e,


let = y dx = dy


Therefore, can be written as


Formula Used: = log |x +|+ C


Since is of the form with change in variable.


= log |y +|+ C


= log |y +|+ C


Since , x + = y and dx = dy


= log |(x + ) +|+C


Therefore,


= log |x + +|+C


Hence ,


= - = log |x + +|+C