What is the integral of arctan(x)?

1 Answer
Aug 12, 2018

inttan^(-1)(x)dx=xtan^(-1)(x)-1/2ln(1+x^2)+C, C in RR

Explanation:

I=inttan^(-1)(x)dx

Using integration by parts :

f(x)=tan^(-1)(x), f'(x)=1/(1+x^2)

g'(x)=1, g(x)=x

I=xtan^(-1)(x)-intx/(1+x^2)dx

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

Let u=1+x^2

du=2xdx

I=xtan^(-1)(x)-1/2int1/udu

=xtan^(-1)(x)-1/2ln(|u|)

=xtan^(-1)(x)-1/2ln(1+x^2)+C

\0/ Here's our answer !