How do you find the integral of #int 1/(4+(x-1)^2)dx#?

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Answer 1 · 2018-04-09T13:55:24

#intdx/(4+(x-1)^2)=1/2arctan((x-1)/2)+C#

This somewhat resembles #1/(a^2+x^2)# but some more work is needed.

#u=x-1#
#du=dx#

We have

#int(du)/(4+u^2)du#

This is in a better form now.

In general, #intdx/(a^2+x^2)=1/aarctan(x/a)+C#

So,

#int(du)/(4+u^2)du=1/2arctan(u/2)+C#

Rewriting in terms of #x# yields

#intdx/(4+(x-1)^2)=1/2arctan((x-1)/2)+C#