Question: The Problem
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[![enter image description here](https://i.stack.imgur.com/8xWjD.jpg)](https://i.stack.imgur.com/8xWjD.jpg)

My Work
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We know that:

$\frac{Q'(z)}{Q(z)}=\sum\limits\_{k=1}^{n}\frac{1}{(z-\alpha\_k)}$

Thus

$\frac{P(z)}{Q(z)}=\sum\limits\_{k=1}^{n}\frac{P(z)}{Q'(z)(z-\alpha\_k)}$

This is about as far as I have traveled any ideas?

Possible Ideas
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I get the feeling that I should use : 

[![enter image description here](https://i.stack.imgur.com/0KH6d.jpg)](https://i.stack.imgur.com/0KH6d.jpg)

But I am not sure of it's relevancy.

Answer: 
we have,$$(\ln(|x\ln x |))'= \frac{\ln(x)+1}{x\cdot \ln(x)}$$
Hence

$$\color{blue}{\int\_{e}^{e^2} \frac{\ln(x)+1}{x\cdot \ln(x)}\,\mathrm dx = \left[\ln(x\ln x )\right]\_e^{e^2}, = \ln 2 +1}$$

**OR you may see that,** 
$$\int\_{e}^{e^2} \frac{\ln(x)+1}{x\cdot \ln(x)}\,\mathrm dx = \int\_{e}^{e^2} \frac{\mathrm d(x\cdot\ln(x))}{x\cdot \ln(x)}\,$$
Setting $u= x\ln x$ yiedls
$$\int\_{e}^{e^2} \frac{\ln(x)+1}{x\cdot \ln(x)}\,\mathrm dx = \int\_{e}^{e^2} \frac{\mathrm d(x\cdot\ln(x))}{x\cdot \ln(x)}\,= \int\_{e}^{2e^2} \frac{\mathrm d u}{u}\,= \ln (2e^2)-\ln e=\ln 2+1$$
**hint**: Let $u = \ln x$. Can you finish it?