How do you integrate #int tan^-1x# by integration by parts method?

1 Answer
Narad T.
Oct 31, 2016

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

Explanation:

The integration by parts is #intu'v=uv-intuv'#
Here we have #u'=1##=>##u=x#
and #v=arctanx##=>##v'=1/(1+x^2)#
If #v=arctanx##=>##x=tanv#
So by differentiating, #1=(1/cos^2v)((dv)/dx)#
#v'=(dv)/dx=cos^2v=1/(1+x^2)#

The integration
#intarctanxdx=xarctanx-int(xdx)/(1+x^2)#
#int(xdx)/(1+x^2)=(1/2)int(2xdx)/(1+x^2)#
Let #u=1+x^2##=>##du=2xdx#
#int(2xdx)/(1+x^2)=int(du)/u=lnu#
#int(xdx)/(1+x^2)=1/2ln(1+x^2)#

And finally #intarctanxdx=xarctanx-ln(1+x^2)/2+C#

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