How do you evaluate the integral #int arctanxdx# from 0 to 2 if it converges?

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Answer 1 · 2017-03-14T16:11:18

First integrate #arctanx# without the bounds:

#I=intarctanxdx#

#{(u=arctanx,=>,du=dx/(1+x^2)),(dv=dx,=>,v=x):}#

Then:

#I=xarctanx-intx/(1+x^2)dx#

Which can be integrated using #t=1+x^2# and the natural logarithm integral:

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

So we see that:

#int_0^2arctanxdx=[xarctanx-1/2ln(1+x^2)]_0^2#

#=2arctan2-1/2ln5-(0arctan0-1/2ln1)#

#=2arctan2-lnsqrt5#

#approx1.4096#