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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.14
  • Last Updated : 28 Mar, 2021

Question 1. Evaluate ∫1/ a2-b2x2 dx

Solution:

Let us assume I = ∫ 1/ (a2-b2x2)dx

take 1/b2 common from above eqation

= 1/b2 ∫ 1/ (a2/b2-x2) dx

= 1/b2 ∫ 1/ (a/b)2-x2) dx



Integrate the above eq then, we get

= 1/b2 1/ 2(a/b) log|(a/b)+x/ (a/b)-x| + c [since ∫1/ a2-x2 dx = 1/2a log|x+a/x-a| + c]

= 1/2ab log|a+bx/ a-bx| + c

Hence, I = 1/2ab log |a+bx/ a-bx| + c

Question 2. Evaluate ∫ 1/ a2x2-b2 dx

Solution:

Let us assume I = ∫ 1/ a2x2-b2 dx

take 1/a2 common from above equation

= 1/a2 ∫ 1/ x2-(b2/a2) dx



= 1/a2 ∫ 1/ x2-(b/a)2 dx

Integrate the above eq then, we get

= (1/a2) 1/(2b/a) log|x-(b/a)/x+(b/a)| + c [since ∫1/ x2-a2 dx = 1/2a log|x-a/x+a| + c]

= 1/2ab log|ax-b/ax+b| + c

Hence, I = 1/2ab log|ax-b/ax+b| + c

Question 3. Evaluate ∫ 1/ a2x2+b2 dx

Solution:

Let us assume I = ∫ 1/ a2x2+b2 dx

take 1/a2 common from above equation

= 1/a2 ∫ 1/ x2+(b2/a2) dx

= 1/a2 ∫ 1/ x2+(b/a)2 dx

Integrate the above eq then, we get

= (1/a2) 1/(b/a)tan-1[x/(b/a)] + c [since ∫1/ x2+a2 dx = 1/a tan-1(x/a) + c]

= 1/ab tan-1(ax/b) + c

Hence, I = 1/ab tan-1(ax/b) + c

Question 4. Evaluate ∫ x2-1/ x2+4 dx

Solution:

Let us assume I = ∫ x2-1/ x2+4 dx

We can write the above eq as below,

= ∫ x2-1+4-4/ x2+4 dx

= ∫ (x2+4)-4-1/ x2+4 dx

= ∫ (x2+4)-5/ x2+4 dx



= ∫ (x2+4)/ x2+4 dx – ∫ 5/ x2+4 dx

= ∫ dx – 5∫ 1/ x2+(2)2 dx

Integrate the above eq then, we get

= x – 5/2 tan-1(x/2) +c [since ∫1/ x2+a2 dx = 1/a tan-1(x/a) + c]

Hence I = x – 5/2 tan-1(x/2) +c

Question 5. Evaluate ∫ 1/ √1+4x2 dx

Solution:

Let us assume I = ∫ 1/ √1+4x2 dx

= ∫ 1/ √1+(2x)2 dx (i)

Let 2x = t

2dx = dt

Put the above value in eq (i)

=1/2 ∫ 1/ √1+t2 dt

Integrate the above eq then, we get

=1/2 log|t+√t2+1| + c [since ∫1/ √x2+a2 dx = log|x+√x2+a2| +c]

=1/2 log|2x+√(2x)2+1| + c

Hence, I =1/2 log|2x+√4x2+1| + c

Question 6. Evaluate ∫1/ √a2+b2x2 dx

Solution:

Let us assume I = ∫1/ √a2+b2x2 dx

= ∫1/ √a2+(bx)2 dx (i)

Let bx = t

bdx = dt

dx = dt/b

Put the above value in eq (i)

= 1/b ∫1/ √a2+(t)2 dt

Integrate the above eq then, we get

= 1/b log|t+√a2+t2|+ c [since ∫1/ √a2+x2 dx = log|x+√a2+x2| + c]

= 1/b log|bx+√a2+(bx)2|+ c

Hence, I = 1/b log|bx+√a2+b2x2|+ c

Question 7. Evaluate ∫1/ √a2-b2x2 dx

Solution:

Let us assume I = ∫1/ √a2-b2x2 dx

= ∫1/ √a2-(bx)2 dx (i)

Let bx = t

bdx = dt

dx = dt/b

Put the above value in eq (i)

= 1/b ∫1/ √a2-(t)2 dt

Integrate the above eq then, we get

= 1/b sin-1(t/a)+ c [since ∫1/ √a2-x2 dx = sin-1 (x/a)+ c]

Hence, I = 1/b sin-1(bx/a)+ c

Question 8. Evaluate ∫1/ √(2-x)2+1 dx

Solution:

Let us assume I = ∫1/ √(2-x)2+1 dx (i)

Let 2-x=t

-dx = dt

Put the above value in eq (i)

= -∫1/ √(t)2+(1)2 dt

Integrate the above eq then, we get

= – log |t+√(t)2+1| + c [since ∫1/ √x2+a2 dx = log|x+√x2+a2| +c]

= – log |(2-x)+√(2-x)2+1| + c

Hence, I = – log |(2-x)+√(2-x)2+1| + c

Question 9. Evaluate ∫1/ √(2-x)2-1 dx

Solution:

Let us assume I = ∫1/ √(2-x)2-1 dx (i)

Let 2-x=t

-dx = dt

Put the above value in eq (i)

= -∫1/ √(t)2-(1)2 dt

Integrate the above eq then, we get

= – log |t+√(t)2-1| + c [since ∫1/ √x2-a2 dx = log|x+√x2-a2| +c]

= – log |(2-x)+√(2-x)2-1| + c

Hence, I = – log |(2-x)+√(2-x)2-1| + c

Question 10. Evaluate ∫ x4+1/ x2+1 dx

Solution:

Let us assume I = ∫ x4+1/ x2+1 dx

= ∫ (x2)2+(1)2/ x2+1 dx

= ∫ (x2+1)2-2x2/ x2+1 dx [a2 +b2 = (a+b)2-2ab]

= ∫ (x2+1)2/ x2+1 dx -∫ 2x2/ x2+1 dx

= ∫ (x2+1) dx – ∫ (2x2+2-2/ x2+1) dx

= ∫ (x2+1) dx – ∫ 2(x2+1)/ x2+1 dx +∫ 2/ x2+1 dx

= ∫ (x2+1) dx – ∫ 2 dx + 2∫ 1/ x2+1 dx

Integrate the above eq then, we get

= x3/3 + x – 2x + 2tan-1(x) + c [since ∫1/ x2+a2 dx = 1/a tan-1(x/a) + c]

= x3/3 – x + 2tan-1(x) + c

Hence, I = x3/3 – x + 2tan-1(x) + c

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