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#