Integration by Partial Fractions, using $\sqrt2$ Integrate $$\int\frac{10x^2 + 13x + 9}{(x^2 - 2)(2x + 1)^2}\mathrm dx$$
The difficulty here is in $x^2 - 2$. In order to determine the coefficients A to D, $\sqrt2$ must be invoked, in factorizing. Firstly, put $x = \sqrt 2$, next $-\sqrt 2$, $-0.5$.
The coefficients come out as fractions, which is unusual.
 A: $$\frac{10x^2+13x+9}{(x^2-2)(2x+1)^2}=\frac{x}{49(x^2-2)}+\frac{157}{49(x^2-2)}-\frac{2}{49(2x+1)}-\frac{20}{7(2x+1)^2}$$
For $\frac{157}{49(x^2-2)}$ only you need to use $\sqrt2$, but I think it's not so hard.
A: Note that the coefficient of $A$ of the term
$\frac{1}{x-\sqrt{2}}$ in the partial fraction expansion of $$f(x)=\frac{10x^2 + 13x + 9}{(x^2 - 2)(2x + 1)^2}$$ can be determined by  evaluating the limit
$$A=\lim_{x\to \sqrt{2}} \left((x-\sqrt{2})\cdot\frac{(10x^2 + 13x + 9)}{ (x^2 - 2)(2x + 1)^2}\right)=
\left.\frac{10x^2 + 13x + 9}{ (x +\sqrt{2})(2x + 1)^2}\right|_{x=\sqrt{2}}=\frac{2+157\sqrt{2}}{196}.$$
Similarly for the  coefficient of $B$ of the term $\frac{1}{x+\sqrt{2}}$ we get $\frac{2-157\sqrt{2}}{196}$.
Hence
$$\frac{A}{x-\sqrt{2}}+\frac{B}{x+\sqrt{2}}=\frac{(A+B)x+\sqrt{2}(A-B)}{x^2-2}=\frac{x+157}{49(x^2-2)}.$$
So far we have 
$$f(x)=\frac{x+157}{49(x^2-2)}+\frac{C}{2x+1}+\frac{D}{(2x+1)^2}$$
and it remains to find the coefficients of $C$ and $D$. They should be
$C=-2/49$ and $D=-20/7$.
P.S. According to your given answer there is a little typo in the integrand (replace $(x^2-2)$ with $(x-2)$). The the integral is easier:
\begin{align*}
\int \frac{10x^2 + 13x + 9}{(x - 2)(2x + 1)^2}\,dx=
&\int\left(\frac{3}{x-2}-\frac{1}{2x+1} -\frac{2}{(2x+1)^2}\right)\,dx\\
\\&=\ln\left(\frac{(x - 2)^3}{\sqrt{2x + 1}}\right) - \frac{1}{2x + 1}+c.
\end{align*}
