Calculate $I= \int_{1}^{e}\frac{(1+\ln x)x}{(1+x\ln x)^2}dx$

Question

Please help me solve this: (level = high school) $$ \int_{1}^{e}\frac{(1+\ln x)x}{(1+x\ln x)^2}\,dx $$

Thanks

Answer 1

If you book is correct, basically the only thing you can do is integrating by parts, noticing that the derivative of $\Big(1+x\log(x)\Big)$ is $\Big(1+\log(x)\Big)$.

Then, setting $u=x$,$v'=\frac{1+\ln x}{(1+x\ln x)^2}$, you have $u'=1$,$v=-\frac{1}{1+x\ln x}$ and then $$\int\frac{(1+\ln x)x}{(1+x\ln x)^2}\,dx=-\frac{x}{1+x\ln x}+\int\frac{dx}{1+x\ln x}$$ and the problem is, again, that there is no closed form for the last integral.

So, confirming what already said by other participants, either the book is wrong or you must go through numerical evaluations.

What I think is that the problem is instead $$\int_{1}^{e}\frac{(1+\ln x)}{(1+x\ln x)^2}\,dx=\frac{e}{1+e}$$