<s>[INST] Find the integral $\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$ $$\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$$$=2\int \frac{x^3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx+3\int \frac{x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
For these two integrals, I have tried Euler substitutions and various types of substitutions of irrational functions, but it seems that those substitutions don't simplify the integral. 
What substitution is useful for this type of integrals?
 [/INST] Hint:
Let
$$I=\int \frac{2x^3+3x^2}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx$$
Then, we have
\begin{align*}
I&=\int \frac{2x^3+x^2-3x+2x^2+x-3+2x+3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx\\
&=\int\frac{x+1}{\sqrt{x^2+2x-3}}\mathrm d x+\int\frac{2x+3}{(2x^2+x-3)\sqrt{x^2+2x-3}}\mathrm dx\\
&=\int\frac{x+1}{\sqrt{x^2+2x-3}}\mathrm d x+\int\frac{2x+3}{(2x+3)(x-1)\sqrt{x^2+2x-3}}\mathrm dx\\
&=\int\frac{x+1}{\sqrt{x^2+2x-3}}\mathrm d x+\int\frac{1}{(x-1)\sqrt{x^2+2x-3}}\mathrm dx\\
\end{align*}
Now, we can make $x+1=2\sec t$
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