How do you integrate $int arctanx$ by integration by parts method?

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Answer 1 · 2016-09-11T01:46:03

$xarctan(x)-1/2ln(1+x^2)+C$

We have:

$intarctan(x)dx$

And we will use the integration by parts formula:

$intudv=uv-intvdu$

So, we set $intudv=intarctan(x)dx$ , so we let:

${:(u=arctan(x)," "" ",dv=dx),(" "" "darr," "" "," "darr),(du=1/(1+x^2)dx," "" ",v=x):}$

Thus:

$intarctan(x)dx=xarctan(x)-intx/(1+x^2)dx$

Solving the second integral:

$=xarctan(x)-1/2int(2x)/(1+x^2)dx$

Let $u=1+x^2$ so $du=2xdx$ :

$=xarctan(x)-1/2int(du)/u$

$=xarctan(x)-1/2lnabsu$

$=xarctan(x)-1/2ln(1+x^2)+C$

How do you integrate int arctanxint arctanx by integration by parts method?