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

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Answer 1 · 2018-06-02T12:56:50

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

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/u#du

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#

Answer 2 · 2018-06-02T12:58:28

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

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#