Class 12 RD Sharma Mathematics Solutions Chapter 19 Indefinite Integrals Exercise 19.14 (original) (raw)

Last Updated : 23 Jul, 2025

Indefinite integrals are a fundamental concept in calculus, representing the antiderivative of a function. They are used to find the area under curves, volumes of solids, and solve various problems in physics, engineering, and economics. In this exercise, we will focus on evaluating indefinite integrals using various techniques such as substitution, integration by parts, and integration by partial fractions.

Important Formulas and Concepts: Indefinite Integrals

**Integration by Substitutions

- ∫f(g(x))g'(x) dx = ∫f(u) du, where u = g(x)

**Integration by Parts

- ∫u dv = uv - ∫v du

**Integration by Partial Fractions

- ∫(px+q)/(x^2+ax+b) dx = ∫(A/(x+c) + B/(x+d)) dx

**Standard Integrals

- ∫e^x dx = e^x + C

- ∫sin(x) dx = -cos(x) + C

- ∫cos(x) dx = sin(x) + C

Class 12 RD Sharma Mathematics Solutions - Exercise 19.14

Question 1. Evaluate ∫1/ a2-b2x2 dx

**Solution:

Let us assume I = ∫ 1/ (a2-b2x2)dx

take 1/b2 common from above equation

= 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

Practice Questions on Indefinite Integrals

**Question 1. ∫ (x² + 1)/(x⁴ + 1) dx

**Question 2. ∫ (x² - x + 1)/(x³ + 1) dx

**Question 3. ∫ (x³ + x)/(x⁴ + 1) dx

**Question 4. ∫ (x² + 2x + 2)/(x³ + x² + x + 1) dx

**Question 5. ∫ (2x + 1)/(x² + x + 1) dx

**Question 6. ∫ (x⁴ + 1)/(x² + 1) dx

**Question 7. ∫ (x³ + x²)/(x⁴ + x² + 1) dx

**Question 8. ∫ (x² + 1)/(x³ + x) dx

**Question 9. ∫ (2x³ + 3x² + 2x + 1)/(x³ + x² + x + 1) dx

**Question 10. ∫ (x³ + 2x² + 2x + 1)/(x⁴ + x³ + x² + x) dx

**Read More:

Conclusion

Exercise 19.14 in RD Sharma's Class 12 Chapter on Indefinite Integrals focuses on integrating rational functions. These are integrals where the integrand is a fraction of polynomials. The key techniques used in this exercise include:

1. Partial fraction decomposition

2. Substitution method

3. Long division of polynomials

4. Recognizing standard integral forms

The problems typically involve breaking down complex fractions into simpler ones, which can then be integrated using standard integral formulas or by applying substitution.