# RD Sharma Class 12 Ex 19.10 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.10 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.10 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.10 Solutions Chapter 19 Indefinite Integrals

### Question 1. Evaluate ∫ x2 √x + 2 dx

Solution:

Let, I = ∫ x2 √x + 2 dx (i)

Substituting x + 2 = t, x= t – 2

dx = dt

Substitute the above value in eq (i)

= ∫ (t – 2)2 √t dt

= ∫ (t2 + 4 – 4t) t1/2 dt

= ∫(t5/2 + 4t1/2 – 4t3/2) dt

Integrate the above eq then, we get

= t7/2/(7/2) + 4t3/2/(3/2) – 4t5/2/(5/2) + c

=(2/7) t7/2 + (8/3) t3/2 – (8/5) t5/2 + c

Now, put the value of t in above eq

=(2/7) (x+2)7/2 + (8/3) (x+2)3/2 – (8/5) (x+2)5/2 + c

Hence, I =(2/7) (x+2)7/2 + (8/3) (x+2)3/2 – (8/5) (x+2)5/2 + c

### Question 2. Integrate ∫ x2/(√x-1) dx

Solution:

Let, I = ∫ x2/(√x-1) dx (i)

Put, x-1 = t, so the value of x=t+1

dx = dt

Put the above value in eq (i)

= ∫ (t+1)2/√t dt

On solving the above eq, we get

= ∫ (t2 + 1 + 2t)/√t dt

= ∫ t3/2 + t-1/2 + 2t-1/2 dt

Integrate the above eq then, we get

= (2/5)t5/2 + 2t1/2 + (4/3)t3/2 + c

= (6t5/2 + 30t1/2 + 20t3/2)/ 15 + c

= (2/15)t1/2 (3t2 + 15 + 10t) + c

= (2/15)(x -1)1/2 (3(x -1)2 + 15 + 10(x -1)) + c

= (2/15)(x -1)1/2 (3(x+ 1 – 2x) + 15 + 10x -10)) + c

= (2/15)(x -1)1/2 (3x2 + 3 – 6x + 15 + 10x -10)) + c

= (2/15)(√x -1) (3x2 + 4x + 8) + c

Hence, I = = (2/15)(√x -1) (3x2 + 4x + 8) + c

### Question 3. Integrate ∫ x2/(√3x + 4) dx

Solution:

Let, I = ∫ x2/(√3x+4) dx (i)

Put, 3x + 4 = t, so the value of x = (t – 4)/3

dx = dt/3

Put the above value in eq (i)

= ∫ ((t – 4)/3)2/ √t dt/3

= (1/3) ∫ (t2 + 16 – 8t)/ 9√t dt

= (1/27) ∫ (t2 + 16 – 8t)/√t dt

= (1/27) ∫ (t3/2 + 16t-1/2 – 8t1/2) dt

Integrate the above eq then, we get

= (1/27) [(2/5)t5/2 – (16/3)t3/2 + 32t1/2]+ c

Now put the value of t in above eq

= (1/27) [(2/5)(3x + 4)5/2 – (16/3)(3x + 4)3/2 + 32(3x + 4)1/2]+ c

Hence, I =(1/27) [(2/5)(3x + 4)5/2 – (16/3)(3x + 4)3/2 + 32(3x + 4)1/2]+ c

### Question 4. Integrate ∫ (2x-1)/ (x-1)2 dx

Solution:

Let, I = ∫ (2x-1)/ (x-1)2 dx

Substituting x-1 = t and dx = dt, we get

= ∫ 2(t + 1)-1 / t2 dt

= ∫ (2t + 2 – 1)/ t2 dt

= ∫ (2t +1) / t2 dt

= ∫ 2t/t2 +1/ t2 dt

= 2∫ 1/t dt + ∫ t-2 dt

Integrate the above eq then, we get

= 2log |t| – t-1 + c

Put the value of t in above eq

= 2log |x-1| – 1/(x-1) + c

Hence, I = 2log |x-1| – 1/(x-1) + c

### Question 5. Integrate ∫(2x2 + 3) √x +2 dx

Solution:

Let, I = ∫(2x2 + 3) √x +2 dx

Substituting x +2 = t and dx = dt, we get

= ∫ [2(t -2)+ 3] √t dt

= ∫ [2(t+ 4 – 4t) + 3] √t dt

= ∫ [2t2 + 8 – 8t + 3] √t dt

= ∫ [2t5/2 + 11t-1/2 – 8t3/2] dt

Integrate the above eq then, we get

= (4/7)t7/2 + (22/3)t3/2 – (16/5) t5/2 + c

= (4/7)(x+2)7/2 + (22/3)(x+2)3/2 – (16/5)(x+2)5/2 + c

Hence, I = (4/7)(x+2)7/2 + (22/3)(x+2)3/2 – (16/5)(x+2)5/2 + c

### Question 6. Integrate ∫ (x2 + 3x + 1)/ (x+1)2 dx

Solution:

Let, I = ∫ (x2 + 3x + 1)/ (x+1)2 dx

Substituting x + 1 = t and dx = dt, we get

= ∫ [(t – 1)2 + 3(t – 1) + 1]/ t2 dt

= ∫ (t2 + 1 – 2t +3t -3 +1)/ t2 dt

= ∫ (t+ t – 1)/ t2 dt

= ∫ t2/t2 + t/ t2 – 1/t2 dt

= ∫ (1 + 1/t – t-2) dt

Integrate the above eq then, we get

= t + log |t| + 1/t + c

Put the value of t in above eq

= (x +1) + log |x +1| + 1/(x+1) + c

### Question 7. Integrate ∫ x2 / (√1-x) dx

Solution:

Let, I =∫ x2 / (√1-x) dx

Substituting 1- x = t and dx = – dt, we get

= ∫ – (1-t)2 /√t dt

= – ∫(1 + t2 – 2t)/ √t dt

= – ∫ t-1/2 + t3/2 – 2t1/2 dt

Integrate the above eq then, we get

= – 2t1/2 + (2/5)t5/2 – (4/3)t3/2 + c

= – (30t1/2 + 6t5/2 – 20t3/2) / 15 + c

= – 2t1/2 /15(15 + 3t2 – 10t) + c

= (- 2/15) √(1-x) (15 + 3(1 -x)2 – 10(1 -x)) + c

= (- 2/15) √(1-x) (15 + 3(1 + x2 – 2x) – 10 + 10x)) + c

= (- 2/15) √(1-x) (5 + 3 + 3x– 6x + 10x) + c

= (- 2/15) √(1-x) (3x+ 4x + 8) + c

Hence, I = (-2/15) (3x+ 4x + 5) √1-x + c

### Question 8. Integrate ∫ x(1 – x)23 dx

Solution:

Let, I = ∫ x(1 – x)23 dx

Substituting 1- x=t and dx = -dt, we get

= – ∫(1-t)t23 dt

= – ∫ (t23 – t24) dt

= ∫ (t24 – t23) dt

Integrate the above eq then, we get

= t25/25 – t24/24 + c

= (1-x)25/25 – (1-x)24/24 + c

Hence, I = (1-x)25/25 – (1-x)24/24 + c

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.