<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] \begin{align}I=\int_2^6 \frac{\ln(x-1)}{x^2+2x+2}dx\end{align}
1) Clean up the logarithm, Perform the change of variable $y=x-1$,
\begin{align}I=\int_1^5 \frac{\ln x}{x^2+4x+5}dx\end{align}
2) One wants to obtain a symmetric polynomial (i.e: $ax^2+bx+a$). Perform the change of variable $x=\sqrt{5}u$,
\begin{align}I&=\sqrt{5}\int_{\frac{1}{\sqrt{5}}}^{\frac{5}{\sqrt{5}}} \frac{\ln \left(u\sqrt{5}\right)}{5u^2+4\sqrt{5}u+5}du\\
&=\sqrt{5}\int_{\frac{1}{\sqrt{5}}}^{\sqrt{5}} \frac{\ln u}{5u^2+4\sqrt{5}u+5}du+\sqrt{5}\ln\left( \sqrt{5}\right)\int_{\frac{1}{\sqrt{5}}}^{\sqrt{5}} \frac{1}{5u^2+4\sqrt{5}u+5}du\\
\end{align}
The first integral is $0$ (perform the change of variable $y=\frac{1}{u}$)
\begin{align}I&=\ln\left( \sqrt{5}\right)\left[\arctan\left(\frac{10u+4\sqrt{5}}{2\sqrt{5}}\right)\right]_{\frac{1}{\sqrt{5}}}^{\sqrt{5}}\\
&=\frac{1}{2}\ln 5\left(\arctan 7-\arctan 3\right)\\
&=\boxed{\frac{1}{2}\arctan\left(\frac{2}{11}\right)\ln 5}
\end{align}
NB:
1) 
a) It's often a good idea to clean up the logarithm.
I mean one perform a change of variable to obtain that the logarithm argument be $x$.
b) Every change of variable $y=\frac{a}{x}$, or $y=ax$, doesn't make "dirty" the logarithm, since,
\begin{align}
\ln (ax)&=\ln a+\ln x\\
\ln\left(\frac{a}{x}\right)&=\ln a-\ln x\end{align}
2) Why do we want to obtain symmetric polynomial?
Because, if $\alpha>1$,$a,b,c$ real,
\begin{align}\int_{\frac{1}{\alpha}}^\alpha\frac{\ln x}{ax^2+bx+a}\,dx=0\end{align}
(i assume that the polynomial doesn't have a root in $\left[\frac{1}{\alpha},\alpha\right]$) 
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