$\int \frac{e^{\left(x^2+4 \ln x\right)}-x^3 e^{x^2}}{x-1} d x$ equals to
Answer & explanation
Correct answer: option 4
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 $
$\Rightarrow I =\frac{1}{2}(t-1) e^t+C=\frac{1}{2}\left(x^2-1\right) e^{x^2}+C$