TITLE: Evaluate $\int_1^e\frac{1+x^2\ln x}{x+x^2 \ln x} dx$
QUESTION [8 upvotes]: $$\int_1^e\dfrac{1+x^2\ln x}{x+x^2 \ln x} dx$$

Attempt: 
I have tried substitutions like $\ln x = t$, but they are just not helping.
I end up with : $\displaystyle\int_0^1 \dfrac{1+e^{2t}t}{1+ e^t t} dt $
Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?

REPLY [5 votes]: Let $$I = \int \frac{1+x^2\ln x}{x+x^2\ln x}dx$$
Divide both Numerator and Denominator by $x^2$
so $$I = \int \frac{\frac{1}{x^2}+\ln x}{\frac{1}{x}+\ln x}dx = \int \frac{\bigg(\frac{1}{x}+\ln x\bigg)-\bigg(\frac{1}{x}-\frac{1}{x^2}\bigg)}{\frac{1}{x}+\ln x}dx$$
so $$I = x-\ln \bigg|\frac{1}{x}+\ln x\bigg|+\mathcal{C}.$$