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

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Answer 1 · 2015-10-20T21:12:48

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

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#