Answer
$$\int\frac{1}{\sqrt{x^2+2x+5}}dx=\sinh^{-1}\frac{x+1}{2}+C$$
Work Step by Step
$$A=\int\frac{1}{\sqrt{x^2+2x+5}}dx$$ $$A=\int\frac{1}{\sqrt{(x+1)^2+4}}dx$$
We set $u=x+1$, which means $$du=dx$$
Therefore, $$A=\int\frac{1}{\sqrt{u^2+4}}du$$
Use Formula 34, which states that
$$\int \frac{dx}{\sqrt{a^2+x^2}}=\sinh^{-1}\frac{x}{a}+C$$
for $a=2$ here.
$$A=\sinh^{-1}\frac{u}{2}+C$$ $$A=\sinh^{-1}\frac{x+1}{2}+C$$
