<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] 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EDIT (the last part was corrected $a$ had been replaced by $x$ in $\ln(a+2)$)
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