Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$ 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
 A: First, let's make a change of variables, $x=3 + u$, which maps $(1,3)$ into $(-1,1)$:
$$
   \int_1^3 \frac{\log(x+2)}{x^2+x+15}\mathrm{d} x = \int_{-1}^1 \frac{\log(4+u)}{u^2 + 6 u+ 23}\mathrm{d} u
$$
Now use $\log(4+u) = \log(4) + \log\left(1+\frac{u}{4}\right)$:
$$\begin{eqnarray}
  \int_1^2 \frac{\log(1+2 u)}{2 u^2+7}\mathrm{d} u &=& \int_{-1}^1 \frac{\log(4)}{u^2 + 6 u+ 23}\mathrm{d} u + \int_{-1}^1 \frac{\log\left(1+\frac{u}{4}\right)}{u^2 + 6 u+ 23}\mathrm{d} u
\end{eqnarray}$$
The first integral is trivially evaluated by completing the squares in the denominator: $u^2+6 u+23 = (u+3)^2 + 14$, giving:
$$
 I_0 = \int_{-1}^1 \frac{\log(4)}{u^2 + 6 u+ 23}\mathrm{d} u = \frac{\log(4)}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) \approx 0.121478
$$
This already gives a good approximation to the correct value ($\approx 0.118650$). The second integral can be done expanding logarithm in a series, and integrating term-wise. The first term:
$$
    \Delta_1 = \int_{-1}^1 \frac{u/4}{u^2 + 6 u+ 23}\mathrm{d} u = -\frac{3}{4} \frac{1}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) + \frac{1}{8} \log\left(\frac{5}{3}\right) \approx -0.001868
$$
The second:
$$
  \Delta_2 = -\frac{1}{2} \int_{-1}^1 \frac{(u/4)^2}{u^2 + 6 u+ 23}\mathrm{d} u = -\frac{1}{16} + \frac{5}{32} \frac{1}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) + \frac{3}{32} \log\left(\frac{5}{3}\right) \approx -0.000918
$$
Combining, $I_0 + \Delta_1+\Delta_2 = 0.118692$ which gives a good approximation.
A: 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.
A: It is not really a simple integral (even if nothing special happens in the range $(1,3)$). Sasha gave a fine approximation (+1) let's provide the dilogarithm answer...
Let's start by factoring the denominator $\ x^2+2x+15$ :
The reduced discriminant is $\Delta=1-1\cdot 15=-14\ $ so that that it will have two complex conjugate solutions : $\ a=-1-i\sqrt{14}\ $ and $\ \overline{a}=-1+i\sqrt{14}$
Let's rewrite a little the integral :
$$I:=\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx=\int_1^3\frac{\ln(x+2)}{(x-a)(x-\overline{a})}dx$$
$$I=\frac 1{a-\overline{a}}\int_1^3 \left(\frac {\ln(x+2)}{x-a}-\frac {\ln(x+2)}{x-\overline{a}}\right)dx=\frac {I1-I2}{a-\overline{a}}$$
The (promised!) dilogarithm function looks like :
$$\operatorname{Li}_2(z)=-\int_0^z \frac {\ln(1-t)}t dt$$
Let's rewrite the first part of our integral the same way :
$$I1=\int_1^3 \frac {\ln(x+2)}{x-a} dx=\int_1^3 \frac {\ln(x-a+a+2)}{x-a} dx$$
$$I1=\int_1^3 \frac {\ln((a+2)(\frac{x-a}{a+2}+1)}{x-a} dx=\int_1^3 \frac {\ln(a+2)+\ln(1-\frac{a-x}{a+2})}{x-a} dx$$
$$I1=\left[\ln(x-a)\ln(a+2)\right]_1^3 -\int_1^3 \frac {\ln(1-\frac{a-x}{a+2})}{a-x} dx$$
set $\ \displaystyle t:=\frac{a-x}{a+2}$ (so that $\displaystyle \frac {dt}t=-\frac{dx}{a-x}$) to get :
$$I1=\left[\ln(x-a)\ln(a+2)\right]_1^3 -\int_{\frac{a-1}{a+2}}^{\frac{a-3}{a+2}} \frac {\ln(1-t)}{t} (-dt)=\left[\ln(x-a)\ln(a+2)-\operatorname{Li}_2\left(\frac{a-x}{a+2}\right)\right]_1^3$$
Of course the second part of the integral will be :
$$I2=\int_1^3 \frac {\ln(x+2)}{x-\overline{a}} dx=\left[\ln(x-\overline{a})\ln(\overline{a}+2)-\operatorname{Li}_2\left(\frac{\overline{a}-x}{\overline{a}+2}\right)\right]_1^3$$
So that your integral should be (with $\ a=-1-i\sqrt{14}\ $ and $\ \overline{a}=-1+i\sqrt{14}$) :
$$I=\frac {\left[\ln(x-a)\ln(a+2)-\ln(x-\overline{a})\ln(\overline{a}+2)+\operatorname{Li}_2\left(\frac{\overline{a}-x}{\overline{a}+2}\right)-\operatorname{Li}_2\left(\frac{a-x}{a+2}\right)\right]_1^3}{a-\overline{a}}\approx 0.11865036886767$$
EDIT (the last part was corrected $a$ had been replaced by $x$ in $\ln(a+2)$)
