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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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is there any relation between $p,q$ and $m$ – Aang Aug 9 '12 at 18:08
As non-mathematicians often do, this posting uses "solve" as a catch-all verb. "Evaluate" would be appropriate. One solves problems; one solves equations; one evaluates expressions. – Michael Hardy Aug 9 '12 at 18:12
FWIW: if you depress the quadratic within the logarithm and then perform partial fraction decomposition, you should end up with integrals of the form $$\int\frac{\log(au^2+c)}{u+f}\mathrm du$$ whose integration will involve the dilogarithm. – J. M. Aug 9 '12 at 18:44
@Chris'sister : I frequently see the word "solve" mis-used in just that way not only in this forum, but in homework handed in by students, and elsewhere. I've never seen it in a paper written by a mathematician. – Michael Hardy Aug 9 '12 at 18:49
If $p,q=0$ then well-known partial fraction gives the answer. If $p=0$, $q\neq 0$ then the result contains the polylogarithm function, see*C.integrate-*Calculator.dflt-‌​&f2=ln%28q*x%2Bt%29%2F%28x^2%2Bm*x%2Bn%29&x=10&y=4&f=Integral.integrand_ln%28q*x%‌​2Bt%29%2F%28x^2%2Bm*x%2Bn%29&f3=x&f=Integral.variable_x&a=*FVarOpt.1-**-.***Integra‌​l.rangestart-.*Integral.rangeend---.**Integral.variable--- – vesszabo Aug 9 '12 at 20:00

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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This doesn't solves anything. After evaluating $I'(p)$ you get such a weird expressions that subsequent integration with respect to $p$ is hopeless. You can check it using WA. – Norbert Aug 9 '12 at 19:22
@ Norbert: I've just checked it and you're right. – user 1618033 Aug 9 '12 at 19:51
Clearly, the people who upvoted didn't try to follow through with this approach. ;) – J. M. Aug 10 '12 at 1:32

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 exist – Norbert Aug 9 '12 at 19:24

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