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

1 Answer
Sasha P.
Oct 20, 2015

I = xarctanx - 1/2 ln (x^2+1) + CI=xarctanx−12ln(x2+1)+C

Explanation:

Integration By Parts: int udv = uv - int vdu∫udv=uv−∫vdu

u = arctanx => du = 1/(x^2+1) dxu=arctanx⇒du=1x2+1dx

dv = dx => v=xdv=dx⇒v=x

I = int arctanx dx = xarctanx - int x* 1/(x^2+1)dxI=∫arctanxdx=xarctanx−∫x⋅1x2+1dx

I = xarctanx - int (xdx)/(x^2+1) = xarctanx - int (1/2d(x^2+1))/(x^2+1)I=xarctanx−∫xdxx2+1=xarctanx−∫12d(x2+1)x2+1

I = xarctanx - 1/2 int (d(x^2+1))/(x^2+1)I=xarctanx−12∫d(x2+1)x2+1

I = xarctanx - 1/2 ln (x^2+1) + CI=xarctanx−12ln(x2+1)+C

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