Answer
$$\int \frac{dx}{2+(x-1)^2}=\frac{\sqrt2}{2}\tan^{-1}\Big(\frac{(x-1)\sqrt2}{2}\Big)+C$$
Work Step by Step
$$A=\int \frac{dx}{2+(x-1)^2}$$
We set $a=x-1$, which means $$da=dx$$
Therefore, $$A=\int\frac{da}{2+a^2}=\int\frac{da}{(\sqrt2)^2+a^2}$$
We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$$
So, $$A=\frac{1}{\sqrt2}\tan^{-1}\frac{a}{\sqrt2}+C$$ $$A=\frac{\sqrt2}{2}\tan^{-1}\frac{a\sqrt2}{2}+C$$ $$A=\frac{\sqrt2}{2}\tan^{-1}\Big(\frac{(x-1)\sqrt2}{2}\Big)+C$$
