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I'd like to know if there are some nice ways to solve this type of integral defined as:

$ \int_{a}^{b} \frac{\ln(px^2+qx+t)}{x^2+mx+n} \ dx$

A few days ago I saw such an integral and it seemed pretty tricky since it's rather hard
to figure out where to start from.

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    Really, your teacher expects you to know dilogarithms?2012-08-10

2 Answers 2

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This might be helpful:

Let $I(p)= \int_{a}^{b} \frac{\ln(px^2+qx+t)}{x^2+mx+n} \ dx$

Now take derivative both sides w.r.t. $p$ gives $I'(p)=\int_{a}^{b} \frac{\partial(\frac{\ln(px^2+qx+t)}{x^2+mx+n})}{\partial p} \ dx$ $=\int_{a}^{b} \frac{x^2}{(x^2+mx+n)(px^2+qx+t)}dx$ which we can integrate using some standard substitutions . Then integrate $I'(p)$ w.r.t. $p$ to get $I(p)$.

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    Clearly, the people who upvoted didn't try to follow through with this approach. ;)2012-08-10
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This may also help, though it does not evaluate this integral completely.

$ \int_{a}^{b} \frac{\log(px^2+qx+t)}{x^2+mx+n} \, dx= \int_{a}^{b} \frac{\log(p(|x-x_1|)(|x-x_2|)}{x^2+mx+n} \, dx= $ where $x_1$ and $x_2$ are the roots of $px^2+qx+t$.

$ \int_{a}^{b} \frac{\log(p)+\log|x-x_1| + \log |x-x_2|}{x^2+mx+n} \, dx= \log(p) \int_{a}^{b} \frac{1}{x^2+mx+n}\, dx+\int_{a}^{b} \frac{\log|x-x_1|}{x^2+mx+n}\, dx+\int_{a}^{b} \frac{\log |x-x_2|}{x^2+mx+n}\, dx $

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    I think you should write $|x-x_1|$ , $|x-x_2|$ instead of $x-x_1$, $x-x_2$. Also you approach doesn't work when this roots doesn't exist2012-08-09