TITLE: Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$
QUESTION [11 upvotes]: Could you please give me a hint on how to compute:
$$
\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx
$$
Thank you for your help

REPLY [7 votes]: This was supposed to be a comment on Raymond's answer, but it got too long. I started with trying to obtain Mayrand's fine expression from the dilogarithmic mess one might obtain through Mathematica or Raymond's route, but wound up with a satisfactorily simple expression.
We start from a version of Raymond's answer with the "elementary portion" already simplified:
$\begin{split}
\frac1{2\sqrt{14}}&\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)+\\
&\frac{i}{2\sqrt{14}}\left(\mathrm{Li}_2\left(\frac{2+i\sqrt{14}}{3}\right)+\mathrm{Li}_2\left(\frac{4-i\sqrt{14}}{5}\right)-\right.\\
&\left.\left(\mathrm{Li}_2\left(\frac{2-i\sqrt{14}}{3}\right)+\mathrm{Li}_2\left(\frac{4+i\sqrt{14}}{5}\right)\right)\right)\end{split}$
I grouped the terms in this way, since this allows the easy application of Landen's identity (see this paper for a survey of the various algebraic dilogarithm identities):
$$\mathrm{Li}_2(x)+\mathrm{Li}_2\left(\frac{x}{x-1}\right)=-\frac12\left(\log(1-x)\right)^2$$
Now, we have the relations
$$\frac{\frac{2\pm i\sqrt{14}}{3}}{\frac{2\pm i\sqrt{14}}{3}-1}=\frac{4\mp i\sqrt{14}}{5}$$
which when used with Landen's identity yields
$$\begin{split}
&\frac1{2\sqrt{14}}\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)+\\
&\frac{i}{2\sqrt{14}}\left(\left(-\frac12\left(\log\left(1-\frac{2+i\sqrt{14}}{3}\right)\right)^2\right)-\left(-\frac12\left(\log\left(1-\frac{2-i\sqrt{14}}{3}\right)\right)^2\right)\right)
\end{split}$$
which, after a few more algebraic manipulations, finally yields
$$\frac1{2\sqrt{14}}\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)-\frac{\arctan\sqrt{14}\log\frac53}{2\sqrt{14}}=\color{blue}{\frac1{2\sqrt{14}}\log\,15\;\arctan\frac{\sqrt{14}}{11}}$$
which is even simpler than Mayrand's original result.