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.
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*}