RD Sharma Class 12 Ex 19.15 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.15 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 19.15 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

RD Sharma Class 12 Ex 19.15 Solutions Chapter 19 Indefinite Integrals

Question 1. Evaluate ∫ 1/(4x² + 12x + 5) dx

Solution:

Let I = ∫ 1/(4x² + 12x + 5) dx

by taking 1/4 common from the above eq

= 1/4 ∫ 1/ x² + 3x + 5/4 dx

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

= 1/4 ∫ 1/ (x + 3/2)² – 1 dx (i)

put (x+ 3/2) = t

dx = dt

put the above value in eq. (i)

= 1/4 ∫ 1/ t² – (1)² dt

Integrate the above eq. then, we get

= 1/4 × 1/2×(1) log |t-1/t+1| +c [since, ∫1/x² – a² dx = 1/2a log|x-a/x+a| +c]

put the value of t in the above eq.

= 1/8 log|x+ 3/2 – 1/x+ 3/2 + 1| + c

Hence, I = 1/8 log|2x+1/ 2x+5| + c

Question 2. Evaluate ∫1/x² – 10x + 34 dx

Solution:

Let I = ∫1/x² – 10x + 34 dx

=∫1/x² – 2x × 5 + (5)² – (5)² + 34 dx

=∫1/ (x – 5)² + 9 dx (i)

substituting (x-1) = t

dx = dt

put the above value in eq. (i)

= ∫ 1/ t² + (3)² dt

Integrate the above eq. then, we get

= 1/3 tan^-1 (t/3) + c [Since, ∫ 1/x² + a² dx = 1/a tan^-1 (x/2) + c]

Put the value of t in the above eq.

Hence, I = 1/3 tan^-1 (x-5/ 3) + c

Question 3. Evaluate ∫ 1/ 1-x-x² dx

Solution:

Let I = ∫ 1/ 1-x-x² dx

= ∫ 1/ -(x² – x – 1) dx

adding and subtracting 1/4 in the denominator to make it a perfect square

= ∫ 1/ -(x² – x + 1/4 – 1 – 1/4) dx

= ∫ 1/ -([x² – x + 1/4] – 1 – 1/4) dx

= ∫ 1/ -([x – 1/2]² – 5/4) dx

= ∫ 1/ (5/4 – [x – 1/2]²) dx

= ∫ 1/ ([√5/2]² – [x – 1/2]²) dx

Integrate the above eq. then, we get

= 1/2(√5/2) log|√5/2 + (x-1/2)/ √5/2 – (x-1/2)| + c [since ∫ 1/a² + x² dx = 1/2a log|x+a/x-a| +c]

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

Question 4. Evaluate ∫ 1/2x² – x – 1 dx

Solution:

Let I = ∫ 1/2x² – x – 1 dx

taking 1/2 common from the above eq.

=1/2 ∫ 1/ x² – x/2 – 1/2 dx

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

= 1/2 ∫ 1/ (x – 1/4)² – 9/16 dx

put, x- 1/4 = t

dx = dt

= 1/2 ∫ 1/ t² – (3/4)² dt

Integrate the above eq. then, we get

= (1/2) 1/[2×(3/4)] log|t-(3/4) / t+(3/4)| + c [Since, ∫ 1/x² -a² dx = 1/2a log|x – a/ x+a| + c]

Put the value of t in above eq.

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

Hence, I = 1/3log|x – 1/2x+1| + c

Question 5. Evaluate ∫ dx/x² + 6x +13

Solution:

Let I =∫ dx/x² + 6x +13 (i)

We have x² + 6x +13 = x² + 6x + 3² – 3² +13 =(x + 3)² + 4

Put the above value in eq. (i)

∫ 1/x² + 6x +13 dx = ∫ 1/(x+3)² + 2² dx

put x+3 = t and

dx = dt

= ∫ dt/ t² + 2²

Integrate the above eq. then, we get

= 1/2 tan^-1 t/2 + c

Put the value of t in above eq.

Hence, I = 1/2 tan^-1 x+3/2 + c

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