`inttan^(-1)xdx=….+C`

Options

A. `(1)/(1+x^(2))`
B. `x tan^(-1)x+(1)/(2)log|1+x^(2)|`
C. `x tan^(-1)x+(1)/(2).(tan^(-1)x)/(1+x^(2))`
D. `x tan^(-1)x-(1)/(2)log|1+x^(2)|`

Answer

D

Explanation

Let `l=int tan^(-1)underset("I")(x).underset("II")(1)dx`
On integration by parts, we get
`tan^(-1)x.x-int(1)/(1+x^(2)).x dx = x tan^(-1)-l_(1)" …(i)"`
where, `" "l_(1)=int(x)/(1+x^(2))dx`
Put `1+x^(2)=t rArr x dx=(1)/(2)dt`
`therefore" "l_(1)=(1)/(2)int (1)/(t)dt=(1)/(2)log t=(1)/(2)log(1+x^(2))`
On putting the value of `l_(1)` in Eq. (i), we get
`l=x tan^(-1)x-(1)/(2)log(1+x^(2))+C`