Question

integration of arctanx

Answer 1

Answer:

∫tan−1(x)dx=xtan−1(x)−12ln(1+x2)+C, C∈R

Explanation:

I=∫tan−1(x)dx

Using integration by parts :

f(x)=tan−1(x), f'(x)=11+x2

g'(x)=1, g(x)=x

I=xtan−1(x)−∫x1+x2dx

=xtan−1(x)−12∫2x1+x2dx

Let u=1+x2

du=2xdx

I=xtan−1(x)−12∫1udu

=xtan−1(x)−12ln(|u|)

=xtan−1(x)−12ln(1+x2)+C

\0/ Here's our answer !


Hope it's help you! ! !