How do you find the integral of int arctan(x) dx?

1 Answer
Oct 20, 2015

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

Explanation:

Integration By Parts: int udv = uv - int vdu

u = arctanx => du = 1/(x^2+1) dx

dv = dx => v=x

I = int arctanx dx = xarctanx - int x* 1/(x^2+1)dx

I = xarctanx - int (xdx)/(x^2+1) = xarctanx - int (1/2d(x^2+1))/(x^2+1)

I = xarctanx - 1/2 int (d(x^2+1))/(x^2+1)

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