What's the integral of int arctan(x) dx ?

2 Answers
Jun 2, 2018

xarctanx-ln(x^2+1)/2+C

Explanation:

Problem:intarctanx
Integrate by parts: intfgprime=fg-intfprimeg
f=arctanx,gprime=1
darr
fprime=1/(x^2+1),g=x:
=xarctanx-intx/(x^2+1)dx

Now solving:
intx/(x^2+1)dx
Substitute u=x^2+1->dx=1/(2x)du
=1/2int1/udu

Now solving:
int1/u du
This is a standard integral
=lnu

Plug in solved integrals:
1/2int1/udu
=lnu/2

Undo substitution u=x^2+1:
=ln(x^2+1)/2

Plug in solved integrals:
=xarctanx-intx/(x^2+1)dx
=xarctanx-ln(x^2+1)/2

The problem is solved:
intarctanx
=xarctanx-ln(x^2+1)/2+C

Jun 2, 2018

int arctan(x) dx = xarctan(x)-1/2log(1+x^2)+C

Explanation:

Integration by parts with u=arctan(x) and (dv)/(d x)=1, giving (du)/(d x) = 1/(1+x^2) and v=x.

Thus
int arctan(x) dx = xarctan(x)-int x/(1+x^2) dx

The second term integrates easily as a natural logarithm:
int arctan(x) dx = xarctan(x)-1/2log(1+x^2)+C