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.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.10
CategoryRD Sharma Solutions

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 ∫ (x+ 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

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