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Today I saw a question with an answer that made me rethink of the following question, since it's not the first time I try to find an answer to it. If you look at the answer of Mhenni Benghorbal
here you'll see $2$ interesting integrals, namely: $$ \int _{0 }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} ; \int _{0}^{\infty }\!{\frac {\ln \left( z \right) }{2+{z}^{2}+2\,z}}dz $$ I try to find out if there is a well defined strategy to tackle such integrals. In a more general sense, we have to deal with:

$$ \int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx +p}}{dx} $$

Could you help here? Thanks.

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  • $\begingroup$ Do you know anything about residues and Cauchy formula? $\endgroup$ – Norbert Aug 30 '12 at 10:29
  • $\begingroup$ @ Norbert: of course, but less practice. $\endgroup$ – user 1357113 Aug 30 '12 at 10:30
  • $\begingroup$ @Norbert: thanks. It would be interesting if such integrals may possibly be solved by some real techniques. $\endgroup$ – user 1357113 Aug 30 '12 at 10:53
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Related problems: (I), (II). You can use the partial fraction technique combined with the use the dilogarithm function $\operatorname{Li}_{2}(x)$, which is defined by

$$\operatorname{Li}_{2}(x) = \int_{1}^{x} \frac{\ln(t)}{1-t} \,dt \,.$$

Here is an example,

$$ \int_{a}^{b} \frac{\ln(x)}{cx+d}dx =- \frac{1}{d}\left( \operatorname{Li}_{2}\left( {\frac {c+da}{c}} \right) +\ln \left(a\right) \ln \left( {\frac {c+da}{c}} \right) -\operatorname{Li}_{2} \left( { \frac {c+bd}{c}} \right) -\ln \left( b \right) \ln \left( {\frac {c+ bd}{c}} \right) \right) $$

Note that the above integral is undefined for $$ \left(a < -\frac{c}{d}, -\frac{c}{d} < b \right) $$

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Have you seen http://www.recreatiimatematice.ro/arhiva/articole/RM12011DICU.pdf

For partial response

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