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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.15

• Last Updated : 18 Mar, 2021

### Question 1. Evaluate âˆ« 1/(4x2 + 12x + 5) dx

Solution:

Let I = âˆ« 1/(4x2 + 12x + 5) dx

by taking 1/4 common from the above eq

= 1/4 âˆ« 1/ x2 + 3x + 5/4 dx

= 1/4 âˆ« 1/ x2 + 2x Ã— (3/2)x + (3/2)2 – (3/2)2 + 5/4 dx

= 1/4 âˆ« 1/ (x + 3/2)2 – 1 dx (i)

put (x+ 3/2) = t

dx = dt

put the above value in eq. (i)

= 1/4 âˆ« 1/ t2 – (1)2 dt

Integrate the above eq. then, we get

= 1/4 Ã— 1/2Ã—(1) log |t-1/t+1| +c [since, âˆ«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 âˆ«1/x2 – 10x + 34 dx

Solution:

Let I = âˆ«1/x2 – 10x + 34 dx

=âˆ«1/x2 – 2x Ã— 5 + (5)2 – (5)2 + 34 dx

=âˆ«1/ (x – 5)2 + 9 dx (i)

substituting (x-1) = t

dx = dt

put the above value in eq. (i)

= âˆ« 1/ t2 + (3)2 dt

Integrate the above eq. then, we get

= 1/3 tan-1 (t/3) + c [Since, âˆ« 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 âˆ« 1/ 1-x-x2 dx

Solution:

Let I = âˆ« 1/ 1-x-x2 dx

= âˆ« 1/ -(x2 – x – 1) dx

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

= âˆ« 1/ -(x2 – x + 1/4 – 1 – 1/4) dx

= âˆ« 1/ -([x2 – x + 1/4] – 1 – 1/4) dx

= âˆ« 1/ -([x – 1/2]2 – 5/4) dx

= âˆ« 1/ (5/4 – [x – 1/2]2) dx

= âˆ« 1/ ([âˆš5/2]2 – [x – 1/2]2) dx

Integrate the above eq. then, we get

= 1/2(âˆš5/2) log|âˆš5/2 + (x-1/2)/ âˆš5/2 – (x-1/2)| + c [since âˆ« 1/a2 + x2 dx = 1/2a log|x+a/x-a| +c]

Hence, I = 1/âˆš5 log|âˆš5/2 + (x-1/2) /âˆš5/2 – (x-1/2)| + c

### Question 4. Evaluate âˆ« 1/2x2 – x – 1 dx

Solution:

Let I = âˆ« 1/2x2 – x – 1 dx

taking 1/2 common from the above eq.

=1/2 âˆ« 1/ x2 – x/2 – 1/2 dx

=1/2 âˆ« 1/ x2 – 2x Ã— 1/4 + (1/4)2 – (1/4)2 – 1/2 dx

= 1/2 âˆ« 1/ (x – 1/4)2 – 9/16 dx

put, x- 1/4 = t

dx = dt

= 1/2 âˆ« 1/ t2 – (3/4)2 dt

Integrate the above eq. then, we get

= (1/2) 1/[2Ã—(3/4)] log|t-(3/4) / t+(3/4)| + c [Since, âˆ« 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 âˆ« dx/x2 + 6x +13

Solution:

Let I =âˆ« 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)

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

put x+3 = t and

dx = dt

= âˆ« 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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