Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.10 (original) (raw)

Last Updated : 18 Mar, 2021

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(x2 + 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)2 + 3] √t dt

= ∫ [2(t2 + 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

= ∫ (t2 + 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 + 3x2 - 6x + 10x) + c

= (- 2/15) √(1-x) (3x2 + 4x + 8) + c

Hence, I = (-2/15) (3x2 + 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