The value of  

$\int \frac{x^5+x^4+4x^3+4x^2+4x+4}{{\left(x^2+2\right)}^5}$ is equal to 

$\begin{array}{l}{x}^5+x^4+4x^3+4x^2+4x+4\\ =x^3\left(x^2+2\right)+x^2\left(x^2+2\right)+2x\left(x^2+2\right)\\ +2\left(x^2+2\right)\\ =\left(x^2+2\right)\left(x^3+x^2+2x+2\right)\\ =\left(x^2+2\right)\left(x\left(x^2+2\right)+\left(x^2+2\right)\right)\end{array}$

$={\left(x^2+2\right)}^2(x+2)$. Therefore,

$\begin{array}{l}f\left(x\right)=\frac{x^5+x^4+4x^3+4x^2+4x+4}{{\left(x^2+2\right)}^5}\\ =\frac{x+2}{{\left(x^2+2\right)}^3}=\frac{1}{2}\frac{2x}{{\left(x^2+2\right)}^3}+\frac{2}{{\left(x^2+2\right)}^3}\end{array}$

Now, $\int \frac{\mathrm{d}x}{{\left(x^2+2\right)}^3}=\frac{1}{8}\frac{x}{{\left(x^2+2\right)}^2}+\frac{3}{8}\int \frac{\mathrm{d}x}{{\left(x^2+2\right)}^2}$

                                                                                              $\left(\text{ using }\int \frac{\mathrm{d}y}{{\left(y^2+k^2\right)}^n}\right)$

$\begin{array}{l}\int \frac{\mathrm{d}x}{{\left(x^2+2\right)}^2}=\frac{1}{4}\frac{x}{x^2+2}+\frac{1}{4\sqrt{2}}={\tan }^{-1}\frac{x}{\sqrt{2}}+C\\ \therefore \int f\left(x\right)dx=-\frac{1}{4{\left(x^2+2\right)}^2}+\frac{1}{4}{\left(x^2+2\right)}^2\\ +\frac{3}{4}\left[\frac{1}{4}\frac{x}{x^2+2}+\frac{1}{4\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}\right]+C\end{array}$

$=\frac{x-1}{4{\left(x^2+2\right)}^2}+\frac{3}{16}\frac{x}{x^2+2}+\frac{3}{16\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}+C$.