Evaluate the following integrals:

$=x^{2}+2 x+10=x^{2}+2 x+1-1+10($ add and substract 1$)$

$\left.=\left(x^{2}+1\right)^{2}-1+10=x^{2}+1\right)^{2}+9$

$=\left(x^{2}+1\right)^{2}+3^{2}$

Put $x+1=t$ hence $d x=d t$ and $x=t-1$

$\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x=\int 1 /\left(\left(x^{2}+1\right)^{2}+3^{2}\right) d x$

$=\int \frac{1}{t^{2}+3^{2}} d t$

We have, $\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\mathrm{a}^{2}}=\frac{1}{\mathrm{a}} \log \left(\frac{\mathrm{t}-\mathrm{a}}{\mathrm{t}+\mathrm{a}}\right)+\mathrm{c}$

Here $a=3$

Therefore, $\int \frac{1}{t^{2}+3^{2}} d t=\frac{1}{3} \log \left(\frac{t-3}{t+3}\right)+c$

Put $t=x+1$

$=\frac{1}{3} \log \left(\frac{t-3}{t+3}\right)+c=\frac{1}{3} \log \left(\frac{x+1-3}{x+1+3}\right)+c=\frac{1}{3} \log \left(\frac{x-2}{x+4}\right)+c$