`int(xtan^-1x)/(1+x^2)^(3/2)dx`A. `(x-tan^(-1)x)/(1-x^(2))+C`B. `(x+tan^(-1)x)/(sqrt(1-x^(2)))+C`C. `(x-tan^(-1)x)/(sqrt(1+x^(2)))+C`D. `(x+sqrt(1-x^(2)))/(sqrt(1+x^(2)))+C`

`int(xtan^-1x)/(1+x^2)^(3/2)dx`A. `(x-tan^(-1)x)/(1-x^(2))+C`B. `(x+ta

1.	`int(xtan^-1x)/(1+x^2)^(3/2)dx`A. `(x-tan^(-1)x)/(1-x^(2))+C`B. `(x+tan^(-1)x)/(sqrt(1-x^(2)))+C`C. `(x-tan^(-1)x)/(sqrt(1+x^(2)))+C`D. `(x+sqrt(1-x^(2)))/(sqrt(1+x^(2)))+C`
Answer» Correct Answer - C Let `l=int(x tan^(-1)x)/((1+x^(2))^(3//2))dx` `"Put "x= tan theta` `rArr" "dx=sec^(2) theta d theta` `therefore" "l=int(tan theta tan^(-1)(tan theta))/((1+tan^(2) theta)^(3//2))sec^(2) theta d theta` `rArr" "l=int(theta tan theta sec^(2) theta)/(sec^(3) theta)d theta= int theta sin theta d theta` By using integration by parts, we get `rArr" "l=theta(-cos theta)+int cos theta d theta` `= - theta cos theta + sin theta+C` `rArr" "l=-tan^(-1)x.(1)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+C` `rArr" "l=((x-tan^(-1)x))/(sqrt(1+x^(2)))+C`