$\int \frac{e^{\left(x^2+4 \ln x\right)}-x^3 e^{x^2}}{x-1} d x$ equals to

$\left(\frac{e^{3 \ln x}-e^{\ln x}}{2 x}\right) e^{x^2}+C$ $\frac{(x-1) x e^{x^2}}{2}+C$ $\frac{\left(x^2-1\right)}{2 x} e^{x^2}+C$ none of these

none of these

Let $I =\int \frac{e^{x^2+4 \ln x}-x^3 e^{x^2}}{x-1} d x=\int \frac{e^{x^2} . x^4-x^3 e^{x^2}}{x-1} d x$ $\Rightarrow I =\int x^3 e^{x^2} d x=\frac{1}{2} \int t e^t d t$, where $t=x^2 $ $\Rightarrow I =\frac{1}{2}(t-1) e^t+C=\frac{1}{2}\left(x^2-1\right) e^{x^2}+C$