<s>[INST] 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
 [/INST] 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.
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