<s>[INST] Evaluate $\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx$ 
Evaluate $$I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx$$

I tried the substitution $t=x-1$ which gave $$I=\int_1^5 \frac{\ln(t)}{t^2+4t+5}dt$$
Then I let $u=\ln t$ which gave $$I=\int_0^{\ln 5} \frac{ue^u}{e^{2u}+4e^u+5}du$$
I don't see other useful substitutions and I don't know how to finish
 [/INST] Symmetry works nice in this case.
$$I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx\,\overset{x-2\rightarrow x}=\,\int_0^4 \frac{\color{red}{\ln(1+x)}}{x^2+6x+10}dx$$
With a substitution of
$$x=\frac{4-y}{1+y}\Rightarrow dx= - \frac{5}{(1+y)^2}dy$$
$$\require{cancel} \Rightarrow I= \int_0^4 \ln\left(1+\frac{4-y}{1+y}\right)\frac{\cancel{(1+y)^2}}{\cancel 5(y^2+6y+10)}\frac{\cancel 5}{\cancel{(1+y)^2}}dy\overset{y=x}=\int_0^4  \frac{\color{blue}{\ln\left(\frac{5}{1+x}\right)}}{x^2+6x +10}dx $$Adding this result with the previous integral gives:
$$\require{cancel} 2I=\int_0^4 \frac{\cancel{\color{red}{\ln(1+x)}}+\color{blue}{\ln 5-\cancel{\ln(1+x)}}}{x^2+6x+10}dx$$
$$\Rightarrow I=\frac{\ln 5}{2} \int_0^4 \frac{1}{x^2+6x+10}dx=\frac{\ln 5}{2} \arctan\left(\frac2{11}\right)$$

One can of course do directly $\displaystyle{x=\frac{t+4}{t-1}}$ when the bounds are $2$ and $6$ to arrive at the same result by the same method.
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