$\int \frac{x}{\left(x^2+2x+2\right)\sqrt{x+1}}dx\text{ is equal to }$

$\int \frac{x}{\left(x^2+2x+2\right)\sqrt{x+1}}dx=\int \frac{x}{\left[{\left(x+1\right)}^2+1\right]\sqrt{x+1}}dx$

$\text{ Put }x+1=t^2$

$\begin{array}{l}dx=2tdt\\ \int \frac{\left(t^2-1\right)2tdt}{\left(t^4+1\right)t}\\ =2\int \frac{t^2\left(1-\frac{1}{t^2}\right)dt}{t^2\left(t^2+\frac{1}{t^2}\right)}\\ =2\int \frac{1-\frac{1}{t^2}}{\left(t^2+\frac{1}{t^2}\right)}dt\\ =2\int \frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-(\sqrt{2}{)}^2}\end{array}$

$\text{ Put }t+\frac{1}{t}=k,\left(1-\frac{1}{t^2}\right)dt=dk$

$\begin{array}{l}=2\int \frac{dk}{k^2-(\sqrt{2}{)}^2}\\ =2\frac{1}{2\sqrt{2}}\log \left|\frac{k-\sqrt{2}}{k+\sqrt{2}}\right|+C\\ =\frac{1}{\sqrt{2}}\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C\\ =\frac{1}{\sqrt{2}}\log \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right|+C\\ \end{array}$

$=\frac{1}{\sqrt{2}}\log \left|\frac{x+1-\sqrt{2}\sqrt{x+1}+1}{x+1+\sqrt{2}\sqrt{x+1}+1}\right|+C$