Class 12 RD Sharma Mathematics Solutions Chapter 19 Indefinite Integrals Exercise 19.19 (original) (raw)

Last Updated : 23 Jul, 2025

In Class 12, Chapter 19 of RD Sharma’s Mathematics textbook focuses on Indefinite Integrals. This chapter delves into the fundamental concept of integration in which is crucial for the solving various problems in calculus. Integrals represent the accumulation of quantities and are pivotal in the finding areas, volumes and solutions to the differential equations.

Indefinite Integrals

The Indefinite integrals also known as antiderivatives represent a family of functions whose derivative is the given function. Unlike definite integrals in which compute the area under a curve between the two limits indefinite integrals do not have specified bounds and include an arbitrary constant of the integration usually denoted as C. The process of finding an indefinite integral is essentially reversing differentiation.

**Class 12 RD Sharma Mathematics Solutions - Exercise 19.19

Question 1. ∫ x/(x2 + 3x + 2) dx

**Solution:

Given that I = ∫ x/(x2 + 3x + 2) dx

Let x = m d/dx (x2 + 3x + 2) + n

= m(2x + 3) + n

x = (2m)x + (3λ + n)

On comparing the coefficients of x,

2m = 1

m = 1/2

3m + n = 0

3(1/2) + n = 0

n = -3/2

I = ∫(1/2(2x + 3) - 3/2)/(x2 + 3x + 2) dx

= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx - 3/2 ∫1/(x2 + 3x + 2) dx

= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx - 3/2 ∫1/(x2 + 2x(3/2) + (3/2)2 - (3/2)2 + 2) dx

= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx - 3/2 ∫1/((x + 3/2)2 - (1/2)2) dx

= 1/2log|x2 + 3x + 2| - (3/2) × (1/2 × (1/2))log|(x + 3/2 - 1/2)/(x + 3/2 + 1/2)| + c

As we know that ∫1/(a2 - x2)dx = 1/2a log|(x - a)/(x + a)| + c]

Hence, I = 1/2log|x2 + 3x + 2| - (3/2) × (1/2 × (1/2))log|(x + 1)/(x + 2)| + c

Question 2. ∫(x + 1)/(x2 + x + 3) dx

**Solution:

Given that I = ∫(x + 1)/(x2 + x + 3) dx

Let us considered x + 1 = m d/dx(x2 + x + 3) + n

x + 1 = m(2x + 1) + n

x + 1 = (2m)x + (m + n)

On comparing the co-efficient of x,

2m = 1

m = 1/2

m + n = 1

(1/2) + n = 1

n = 1/2

Now,

I = ∫(1/2(2x + 1) + 1/2)/(x2 + x + 3) dx

=1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/(x2 + 2x(1/2) + (1/2)²-(1/2)²+3) dx

= 1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/((x + 1/2)2 + (11/4)) dx

= 1/2 ∫(2x + 1)/(x2 + x + 1) dx + 1/2 ∫1/((x + 1/2)2 + (√11/2)2) dx

= 1/2 log|x2 + x + 3| + (1/2) * (1/(√11/2)) tan-1((x + 1/2)/(√11/2)) + c

As we know that ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c

Hence, I = 1/2 log|x2 + x + 3| + 1/√11 tan-1((2x + 1)/(√11)) + c

Question 3. ∫ (x - 3)/(x2 + 2x - 4) dx

**Solution:

Given that I = ∫(x - 3)/(x2 + 2x - 4) dx

Let us considered x - 3 = m d/dx (x2 + 2x - 4) + n

= m(2x + 2) + n

x - 3 = (2m)x + (2m + n)

On comparing the coefficients of x,

2m = 1

m = 1/2

2m + n = -3

2(1/2) + n = -3

n = -4

So,

I = ∫(1/2(2x + 2) - 4)/(x2 + 2x - 4) dx

= 1/2 ∫(2x + 2)/(x2 + 2x - 4) dx - 4∫ 1/(x2 + 2x + (1)2 - (1)2 - 4) dx

= 1/2 ∫(2x + 2)/(x2 + 2x - 4) dx - 4∫ 1/((x + 1)2 - (√5)) dx

As we know that ∫1/(x2 - a2) dx = 1/2a log|(x - a)/(x + a)| + c

= 1/2 log⁡|x2 + 2x - 4| - 4 × 1/(2√5) log⁡|(x + 1 - √5)/(x + 1 + √5)| + c

Hence, I = 1/2 log⁡|x2 + 2x - 4| - 2/√5 log⁡|(x + 1 - √5)/(x + 1 + √5)| + c

Question 4. ∫(2x - 3)/(x2 + 6x + 13) dx

**Solution:

Given that I = ∫ (2x - 3)/(x2 + 6x + 13) dx

Let us considered 2x - 3 = m d/dx (x2 + 6x + 13) + n

= m(2x + 6) + n

2x - 3 = (2m)x + (6m + n)

On comparing the co-efficient of x, we get

2m = 2

m = 1

6m + n = -3

6 * 1 + n = -3

n = -9

Now,

= ∫(1 * (2x + 6) - 9)/(x2 + 6x + 13) dx

= ∫(2x + 6)/(x2 + 6x + 13) dx + ∫(-9)/(x2 + 2 * (3) * x + (3)2 - (3)2 + 13) dx

= ∫(2x + 6)/(x2 + 6x + 13) dx -9 ∫1/((x + 3)2 + (2)) dx

= log|(x2 + 6x + 13)| - 9 * (1/2) tan-1((x + 3)/2) + c

Hence, I = log|(x2 + 6x + 13)| - 9 × (1/2) tan-1((x + 3)/2) + c

Question 5. ∫x2/(x2 + 7x + 10) dx

**Solution:

Given that I = ∫x2/(x2 + 7x + 10) dx

= ∫{1 - (7x + 10)/(x2 + 7x + 10)}dx

I = x - ∫(7x + 10)/(x2 + 7x + 10) dx + c1 ........(i)

Let I1 = ∫(7x + 10)/(x2 + 7x + 10) dx

Let 7x + 10 = md/dx (x2 + 7x + 10) + n

= m(2x + 7) + n

7x + 10 = (2m)x + 7m + n

On comparing the coefficients of like powers of x,

7 = 2m

m = 7/2

7m + n = 10

7(7/2) + n = 10

n = -29/2

So, l = ∫(1/6(6x - 4) - 1/3)/(3x2 - 4x + 3) dx

= 1/6 ∫(6x - 4)/(3x2 - 4x + 3) dx - 1/9 ∫1/(x2 - 4/3x + 1) dx

= 1/6 ∫(6x - 4)/(3x2 - 4x + 3) dx - 1/9 ∫1/(x2 - 2x(2/3) + (2/3)2 - (2/3)2 + (2)2) dx

= 1/6 ∫(6x - 4)/(3x2 - 4x + 3) dx - 1/9 ∫1/(x - 2/3)2 + (√5/2)) dx

= 1/6log|(3x2 - 4x + 3) | - ((1/9) × 1/(√5/3)) tan((x - 2/3)/(√5/3)) + c

Hence, I = 1/6log|(3x2 - 4x + 3)| - (√5/15)tan-1((3x - 2)/√5) + c

Question 6. ∫2x/(2 + x - x2) dx

**Solution:

Given that I = ∫2x/(2 + x - x2) dx

Now,

2x = m(d/dx(2 + x + x2)) + n

2x = m(-2x + 1) + n

Now equating the co-efficient of we will get m, n

m = -1,

n = 1

∫2x/(2 + x - x2) dx

= ∫(m(-2x + 1) + n)/(2 + x - x2) dx

= ∫(-1(-2x + 1) + 1)/(2 + x - x2) dx

= ∫(-1(-2x + 1))/(2 + x - x2) dx + 1/(2 + x - x2) dx

= -log⁡|2 + x - x2| + ∫1/(2 + x - x2) dx

= -log⁡|2 + x - x2| - ∫1/(x2 - x - 2) dx

= -log⁡|2 + x - x2 - ∫1/(x2 - x(1/2)(2) + (1/2) - (1/2) - 2) dx

= -log⁡|2 + x - x2| + ∫1/((x - 1/2)2 - (3/2)2) dx

= -log⁡|2 + x - x2| - 1/3 log|((x - 1/2) - (3/2))/((x - 1/2) + (3/2))| + c

Hence, I = -log⁡|2 + x - x2| - 1/3 log⁡|(x - 2)/(x + 1)| + c

Question 7. ∫(1 - 3x)/(3x2 + 4x + 2) dx

**Solution:

Given that I = ∫(1 - 3x)/(3x2 + 4x + 2) dx

Let us considered 1 - 3x = m d/dx (3x2 + 4x + 2) + n

= m(6x + 4) + n - 11 - 3x = (6m)× + (4λ + n)

On comparing the coefficients of l x,

6m = -3

m = -1/2

4m + n = 1

4(-1/2) + n = 1

n = 3

I = ∫(-1/2(6x + 4) + 3)/(3x2 + 4x + 2) dx

= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3∫1/(3x2 + 4x + 2) dx

= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3/3 ∫1/(x2 + 4/3x + 2/3) dx

= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/(x2 + 2x(2/3) + (2/3)2 - (2/3)2 + (2/3) dx

= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + 2/9) dx

= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + (√2/3)2) dx

= -1/2 log|(3x2 + 4x + 2) | + 3/√2tan-1((x + 2/3)/(√2/3)) + c

Hence, I = -1/2 log|(3x2 + 4x + 2)| + 3/√2tan-1((3x + 2)/√2) + c

Question 8. ∫(2x + 5)/(x2 - x - 2) dx

**Solution:

Given that I = ∫(2x + 5)/(x2 - x - 2) dx

Let 2x + 5 = md/dx (x2 - x - 2) + n

= m(2x - 1) + n

2x + 5 = (2m)x - m + n

On comparing the coefficients of x,

2m = 2

m = 1

-m + n = 5

-1 + n = 5

n = 6

So,

I = ∫((2x - 1) + 6)/(x2 - x - 2) dx

= ∫ ((2x - 1))/(x2 - x - 2) dx + 6∫1/(x2 - 2x(1/2) + (1/2)2 - (1/2)2 - 2) dx

= ∫ (2x - 1)/(x2 - x - 2) dx + 6∫1/((x - 1/2)2 - 9/4) dx

= ∫ (2x - 1)/(x2 - x - 2) dx + 6∫1/((x - 1/2)2 - (3/2)2) dx

= log⁡|x2 - x - 2| + 6/2(3/2) log⁡|(x - 1/2 - 3/2)/(x - 1/2 + 3/2)| + c

As we know that ∫1/(x2 - a2) dx = 1/2a log⁡|(x - a)/(x + a)| + c

Hence, I = log⁡|x2 - x - 2| + 2log⁡|(x - 2)/(x + 1)| + c

Question 9. ∫ (ax3 + bx)/(x4 + c2) dx

**Solution:

Given that I = ∫(ax3 + bx)/(x4 + c2) dx

Let us considered ax3 + bx = md/dx (x4 + c2) + n

ax3 + bx = n(4x3) + n

On comparing the coefficients of x,

4m = a

m = a/4

and

n = 0

I = ∫(a/4 (4x3) + bx)/(x4 + c2) dx

= a/4 ∫(4x3)/(x4 + c2) dx + b∫x/((x2)2 + c2) dx

= a/4 ∫(4x3)/(x4 + c2) dx + b/2 ∫2x/((x2)2 + c2) dx

= a/4 log⁡|x4 + c2| + b/2 I1 ........(i)

Now,

I1 = ∫2x/((x2)2 + c2) dx

Put x2 = t

2xdx = dt

I1 = ∫1/((t)2 + c2) dx

= 1/c tan-1⁡(t/c) + c1

l1 = 1/C tan-1⁡(x2/c) + c1 ............(ii)

Now using equation (i) and (ii) we get,

Hence, I = a/4 log|x4 + c2| + b/2c tan(x2/c) + b

Question 10. ∫(x + 2)/(2x2 + 6x + 5) dx

**Solution:

Given that I = ∫(x + 2)/(2x2 + 6x + 5) dx

Let us considered x + 2 = m d/dx (2x2 + 6x + 5) + n

= m(4x + 6) + n

x + 2 = (4m)x + (6m + n)

On comparing the coefficients of x,

So,

4m = 1

m = 1/4

6m + n = 2

6(1/4) + n = 2

n = 1/2

I = ∫(1/4(4x + 6)+1/2)/(2x2 + 6x + 5) dx)

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/2 ∫1/(2x2 + 6x + 5) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 3x + 5/2) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 2x(3/2) + (3/2)2 - (3/2)2 + 5/2) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + 1/4) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + (1/2)2) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 × (1/(1/2))tan-1⁡((x + 3/2)/(1/2)) + c

As we know that ∫1/(x2 + a2) dx = 1/a tan-1⁡(x/a) + c

Hence, I = 1/4 log|2x2 + 6x + 5| + 1/2 tan-1(2x + 3) + c

Question 11. ∫((3sin⁡x - 2)cos⁡x)/(5 - cos2⁡x - 4sin⁡x) dx

**Solution:

Given that I = ∫((3sin⁡x - 2)cos⁡x)/(5 - cos2x - 4sin⁡x) dx

= ∫((3sin⁡x - 2)cos⁡x)/(5 - (1 - sin2⁡x) - 4sin⁡x) dx

= ∫((3sin⁡x - 2)sin⁡x)/(5 - 1 + sin2x - 4sin⁡x) dx)

Now substitute sin⁡x = t in the above equation

cos⁡xdx = dt

So,

I = ∫(3t - 2)/(4 + t2 - 4t) dt

= ∫((3t - 2))/(t2 - 4t + 4) dt

= ∫(3t - 2)/(t - 2)2 dt

Now Integrate partial fractions.

(3t - 2)/((t - 2)2) = A/((t - 2)) + B/((t - 2)2)

= (A(t - 2) + B)/((t - 2)2)

= (At - 2A + B)/((t - 2)2)

3t - 2 = At - 2A + B

On comparing the coefficients, we have, A = 3

and -2A + B = -2

Now, on substituting the value of A = 3 in the above equation,

-2 × 3 + B = -2

-6 + B = -2

B = 6 - 2

B = 4

So, I = ∫(3t - 2)/(t - 2)2dt

= ∫(3t - 2)/(t - 2) dt + ∫4/(t - 2)2 dt

= 3log|t - 2| - 4(1/(t - 2)) + c

= 3log|t - 2| - 4(1/(t - 2)) + c

Now put the value of t = sinx, we have ,

I = 3log|sinx - 2| - 4(1/(sinx - 2)) + c

Question 12. ∫(5x - 2)/(1 + 2x + 3x2) dx

**Solution:

Given that I = ∫(5x - 2)/(1 + 2x + 3x2) dx

Let us considered 5x - 2 = A d/dx (1 + 2x + 3x2) + B

5x - 2 = A(2 + 6x) + B

5x - 2 = 6 × A + 2A + B

On comparing the Co-efficient we have, 6A = 5 and 2A + B = -2

A = 5/6

On substituting the value of A in 2A + B = -2, we n have,

2 * 5/6 + B = -2

10/6 + B = -2

B = -2 - 10/6

B = (-12 - 10)/6

B = (-22)/6

B = (-11)/3

5x - 2 = 5/6(2 + 6x) - 11/3

So, I = ∫(5x - 2)/(1 + 2x + 3x2) dx becomes,

I = ∫[5/6(2 + 6x) - 11/3]/(3x2 + 2x + 1) dx

= 5/6∫(2 + 6x)/(3x2+ 2x + 1) dx - 11/3∫dx/(3x2 + 2x + 1)

= 5/6 log⁡(3x2 + 2x + 1) - 11/(3 × 3)∫dx/(x2 + 2/3x + 1/3) + c

= 5/6 log⁡(3x2 + 2x + 1) - 11/9∫dx/(x2 + 2/3 x + (4/3)2 + 1/3 - (4/3)2) + c

= 5/6 log⁡(3x2 + 2x + 1) - 11/9∫dx/((x + 1/3)2 +(√2/3)2) + c

= 5/6 log⁡(3x2 + 2x + 1) - 11/9 × 1/(√2/3) tan-1(((x + 1/3)/(√2/3))] + C

Hence, I = 5/6 log⁡(3x2 + 2x + 1) - 11/(3√2) tan-1⁡[(3x + 1)/√2] + C

Practice Questions on Indefinite Integrals

1. ∫ (2x + 3)^5 dx

2. ∫ x√(x^2 + 1) dx

3. ∫ (x^2 + 1) / (x^3 + x) dx

4. ∫ tan^2 x dx

5. ∫ x e^x dx

**Read More:

Conclusion

Indefinite integrals serve as a fundamental concept in calculus, representing the reverse process of differentiation. They provide a powerful tool for finding general solutions to differential equations and for calculating areas under curves without specific boundaries. Solving indefinite integrals is a critical skill in calculus, requiring a deep understanding of various techniques and methods such as:

1. Substitution Method: Used when the integrand is a composite function. Let u = g(x), then ∫ f(g(x))g'(x) dx = ∫ f(u) du.

2. Integration by Parts: Used for products of functions. Formula: ∫ u dv = uv - ∫ v du.

3. Partial Fractions: Used for integrating rational functions. Involves breaking complex fractions into simpler terms.

4. Trigonometric Integrals: Special techniques for integrating trigonometric functions, often involving trigonometric identities.

5. Exponential and Logarithmic Integrals: Techniques for integrating expressions involving e^x and ln(x).