Find (ab) if $ \int \frac{15x^2dx}{\sqrt{2x-1}}=\left(ax^2\:+bx\:+b\right)\sqrt{2x-1}+C $ $$
\int \frac{15x^2dx}{\sqrt{2x-1}}=\left(ax^2\:+bx\:+b\right)\sqrt{2x-1}+C
$$
What is the answer written as integer 
$$
225\sqrt{2x-1}+C
$$
(excuse my English)

\begin{eqnarray*}
\int{\frac{15 x^{2}}{\sqrt{2 x-1}} dx} &=& \left(2+2 x+ 3 x^{2} \right) \sqrt{2 x -1} + C
\end{eqnarray*}
which means, $a=3, b=2$ and hence $ab=6$. Is this that you are seeking here?
Note: $y^{2} = 2 x-1$, which means $2 y d y = 2 dx $ or $ \frac{dx}{y}=dy$. The problem then translates to,
\begin{eqnarray*}
\int{\frac{15 x^{2}}{\sqrt{2 x-1}} dx} &=& \int{15 \left(\frac{y^{2}+1}{2} \right)^{2} dy} \\
&=& \frac{y}{4} \left(15+10 y^{2}+ 3 y^{4}\right)  +C\\
&=& \frac{\sqrt{2 x -1} }{4} \left(8+8 x+ 12 x^{2} \right) + C \\
&=&\left(2+2 x+ 3 x^{2} \right) \sqrt{2 x -1} + C
\end{eqnarray*}
