What is #int tan^-1 x dx #?

1 Answer
Feb 20, 2018

#I=tan^-1(x)x-1/2ln(x^2+1)+C#

Explanation:

We want to solve

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

Use integration by parts / partial integration

#intudv=uv-intvdu#

Let #u=tan^-1(x)# and #dv=1dx#

Then #du=1/(x^2+1)dx# and #v=x#

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

Make a substitution #u=x^2+1=>(du)/dx=2x#

#I=tan^-1(x)x-1/2int1/(u)du#

#=tan^-1(x)x-1/2ln(u)+C#

Substitute back #u=x^2+1#

#I=tan^-1(x)x-1/2ln(x^2+1)+C#