Integrate $$ \int_{1}^{\mathrm{e}}\exp\left(\frac{x^{2} - 1}{2}\right) \left[{1 \over x} + x\log\left(x\right)\right]\mathrm{d}x $$
I am unable to understand which property to use. Comment on the technique used and help understand how to proceed with such.
Thanks in advance !.
HINT:
$$\dfrac{d(e^{x^2/2})}{dx}=x\cdot e^{x^2/2}\implies\int x\cdot e^{x^2/2}\ dx= e^{x^2/2}$$
Use Integration by parts
By Parts $$\int \frac{e^{\frac{x^2-1}{2}}}{x}dx=e^{\frac{x^2-1}{2}}lnx-\int e^{\frac{x^2-1}{2}}xlnx\:dx$$
$$\int_{1}^{e}e^{\frac{x^2-1}{2}}\left(\frac{1}{x}+x\log x\right)\,dx = \int_{0}^{1}e^{\frac{e^{2u}-1}{2}}\left(1+ u e^{2u}\right)\,du=\left.u\cdot e^{\frac{e^{2u}-1}{2}}\right|_{0}^{1}.$$
Remark that $$\int e^{\left(\frac{x^{2}-1}{2}\right)} \left(\frac{1}{x}+x\log{x}\right) dx= \int e^{g(x)}[f'(x) + g'(x)f(x)] dx$$ with $f(x)=\log x$ and $g(x)=(\frac{x^{2}-1}{2}).$
So according the the formula $$\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$$
we have
$$\int e^{(\frac{x^{2}-1}{2})}\lgroup \frac{1}{x}+x\log{x}\rgroup dx =f(x)e^{g(x)} = -(\log x)e^{(\frac{x^{2}-1}{2})}+C.$$
Therefore
$$\int_{1}^{e}e^{\frac{x^2-1}{2}}\left(\frac{1}{x}+x\log x\right)\,dx=[-(\log x)e^{(\frac{x^{2}-1}{2})}]_{1}^{e}=e^{(\frac{e^{2}-1}{2})}$$
For the proof of the formula one can see:
Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$
