$$\int\frac{2x^3+3x^2}{\left(2x^2+x-3\right)\sqrt{x^2+2x-3}}\space\text{d}x=$$
Using the Euler subsitution:
$x=\frac{u^2+3}{2u+2}$ and $\text{d}x=\left(\frac{2u}{2u+2}-\frac{2\left(u^2+3\right)}{\left(2u+2\right)^2}\right)\space\text{d}u$.
And when we substitute back we get $u=x+\sqrt{x^2+2x-3}$:
$$\int\frac{\left(u^2+3\right)^2}{2\left(u^2-1\right)^2}\space\text{d}u=\frac{1}{2}\int\frac{\left(u^2+3\right)^2}{\left(u^2-1\right)^2}\space\text{d}u=$$
$$\frac{1}{2}\int\left[\frac{4}{(u+1)^2}+\frac{4}{(u-1)^2}+1\right]\space\text{d}u=$$
$$\frac{1}{2}\left[4\int\frac{1}{(u+1)^2}\space\text{d}u+4\int\frac{1}{(u-1)^2}\space\text{d}u+\int1\space\text{d}u\right]=$$
$$\frac{1}{2}\left[4\int\frac{1}{(u+1)^2}\space\text{d}u+4\int\frac{1}{(u-1)^2}\space\text{d}u+u\right]=$$
Substitute $s=u+1$ and $\text{d}s=\text{d}u$.
And Substitute $p=u-1$ and $\text{d}p=\text{d}u$:
$$\frac{1}{2}\left[4\int\frac{1}{s^2}\space\text{d}s+4\int\frac{1}{p^2}\space\text{d}p+u\right]=$$
$$\frac{1}{2}\left[u-\frac{4}{s}-\frac{4}{p}\right]+\text{C}=$$
$$\frac{1}{2}\left[u-\frac{4}{u+1}-\frac{4}{u-1}\right]+\text{C}=$$
$$\frac{1}{2}\left[x+\sqrt{x^2+2x-3}-\frac{4}{x+\sqrt{x^2+2x-3}+1}-\frac{4}{x+\sqrt{x^2+2x-3}-1}\right]+\text{C}=$$
$$\frac{1}{2}\left[\frac{(3+x)(2x-3)}{\sqrt{(3+x)(x-1)}}\right]+\text{C}=\frac{(3+x)(2x-3)}{2\sqrt{(3+x)(x-1)}}+\text{C}$$