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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.10

• Last Updated : 18 Mar, 2021

### Question 1. Evaluate âˆ« x2 âˆšx + 2 dx

Solution:

Let, I = âˆ« x2 âˆšx + 2 dx (i)

Substituting x + 2 = t, x= t – 2

dx = dt

Substitute the above value in eq (i)

= âˆ« (t – 2)2 âˆšt dt

= âˆ« (t2 + 4 – 4t) t1/2 dt

= âˆ«(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 âˆ« x2/(âˆšx-1) dx

Solution:

Let, I = âˆ« x2/(âˆšx-1) dx (i)

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

dx = dt

Put the above value in eq (i)

= âˆ« (t+1)2/âˆšt dt

On solving the above eq, we get

= âˆ« (t2 + 1 + 2t)/âˆšt dt

= âˆ« 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)(âˆšx -1) (3x2 + 4x + 8) + c

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

### Question 3. Integrate âˆ« x2/(âˆš3x + 4) dx

Solution:

Let, I = âˆ« x2/(âˆš3x+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)

= âˆ« ((t – 4)/3)2/ âˆšt dt/3

= (1/3) âˆ« (t2 + 16 – 8t)/ 9âˆšt dt

= (1/27) âˆ« (t2 + 16 – 8t)/âˆšt dt

= (1/27) âˆ« (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 âˆ« (2x-1)/ (x-1)2 dx

Solution:

Let, I = âˆ« (2x-1)/ (x-1)2 dx

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

= âˆ« 2(t + 1)-1 / t2 dt

= âˆ« (2t + 2 – 1)/ t2 dt

= âˆ« (2t +1) / t2 dt

= âˆ« 2t/t2 +1/ t2 dt

= 2âˆ« 1/t dt + âˆ« 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 âˆ«(2x2 + 3) âˆšx +2 dx

Solution:

Let, I = âˆ«(2x2 + 3) âˆšx +2 dx

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

= âˆ« [2(t -2)2 + 3] âˆšt dt

= âˆ« [2(t2 + 4 – 4t) + 3] âˆšt dt

= âˆ« [2t2 + 8 – 8t + 3] âˆšt dt

= âˆ« [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 âˆ« (x2 + 3x + 1)/ (x+1)2 dx

Solution:

Let, I = âˆ« (x2 + 3x + 1)/ (x+1)2 dx

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

= âˆ« [(t – 1)2 + 3(t – 1) + 1]/ t2 dt

= âˆ« (t2 + 1 – 2t +3t -3 +1)/ t2 dt

= âˆ« (t2 + t – 1)/ t2 dt

= âˆ« t2/t2 + t/ t2 – 1/t2 dt

= âˆ« (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 âˆ« x2 / (âˆš1-x) dx

Solution:

Let, I =âˆ« x2 / (âˆš1-x) dx

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

= âˆ« – (1-t)2 /âˆšt dt

= – âˆ«(1 + t2 – 2t)/ âˆšt dt

= – âˆ« 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) âˆš(1-x) (15 + 3(1 -x)2 – 10(1 -x)) + c

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

= (- 2/15) âˆš(1-x) (5 + 3 + 3x2 – 6x + 10x) + c

= (- 2/15) âˆš(1-x) (3x2 + 4x + 8) + c

Hence, I = (-2/15) (3x2 + 4x + 5) âˆš1-x + c

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

Solution:

Let, I = âˆ« x(1 – x)23 dx

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

= – âˆ«(1-t)t23 dt

= – âˆ« (t23 – t24) dt

= âˆ« (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

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