Evaluate the following integral:
Question:

Evaluate the following integral:

$\int \frac{x}{\left(x^{2}+2 x+2\right) \sqrt{x+1}} d x$

Solution:

assume $x+1=t^{2}$

$\mathrm{d} x=2 \mathrm{tdt}$

$\int \frac{2\left(t^{2}-1\right) d t}{t^{4}+1}$

Dividing by $t^{2}$ in both numerator and denominator

$\int \frac{2\left(1-\frac{1}{t^{2}}\right) d t}{t^{2}+\frac{1}{t^{2}}}$

$\int \frac{2\left(1-\frac{1}{t^{2}}\right) d t}{\left(t+\frac{1}{t}\right)^{2}-2}$

Let $\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)=\mathrm{z}$

$\left(1-\frac{1}{t^{2}}\right) d t=d z$

$\int \frac{2 d z}{z^{2}-2}$

Using identity $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c$

$\frac{1}{\sqrt{2}} \log \left|\frac{z-\sqrt{2}}{z+\sqrt{2}}\right|+c$

Substituting $\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)=\mathrm{z}$

$\frac{1}{\sqrt{2}} \log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+c$

Substituting $\mathrm{t}=\sqrt{\mathrm{x}+1}$

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