`int arctanx dx` Find the indefinite integral

Expert Answers

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Given ,

`int tan^(-1) (x)dx`

By Applying the integration by parts we get this solution


let` u=tan^(-1) (x) =>

`u'= (tan^(-1) (x) )'=1/(x^2+1)`


and `v'=1 =>v =x`

now by Integration by parts ,

`int uv' dx= uv-int u'v dx`

so , now

`int tan^(-1) (x)dx`

= `xtan^(-1) (x) - int 1/(x^2+1)*x dx`

=` xtan^(-1) (x) - int x/(x^2+1)dx`

=` xtan^(-1) (x) - 1/2int 2x/(x^2+1)dx`

let `q=x^2+1`

=> `dq = 2x dx`

so ,

`1/2int 2x/(x^2+1)dx = 1/2int 1/q dq = 1/2ln(q)+C`


 so, now

`int tan^(-1) (x)dx`= ` xtan^(-1) (x) -1/2ln(x^2+1)+C`

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