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

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

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