Evaluating $ \int_{0}^{6}\frac{\log(x)}{1+x^2}\,dx$ I'm trying to understand an answer by @Jack D' Aurizio. 
This is the part I get lost on:
$$ \int_{0}^{6}\frac{\log(x)}{1+x^2}\,dx = -\int_{1/6}^{+\infty}\frac{\log(x)}{1+x^2}\,dx=\int_{0}^{1/6}\frac{\log(x)}{1+x^2}\,dx$$
What is going on here? I think Jack first does the substitution $u=\frac{1}{x}$ but then I'm lost.
Also Jack says,
$$ \int_{0}^{1/6}x^{2k}\log(x)\,dx = -\frac{1+(2k+1)\log 6}{6^{2k+1}(2k+1)^2}$$
Again I don't understand what's happening.
 A: Start with the integral 
$$\begin{align}
I&=\int_0^\infty \frac{\log(x)}{1+x^2}\,dx\\\\
&=\int_0^6 \frac{\log(x)}{1+x^2}\,dx+\int_6^\infty \frac{\log(x)}{1+x^2}\,dx \tag 1
\end{align}$$
Enforcing the substitution $x\to 1/x$ in integrals on the right-hand side of $(1)$ yields
$$\begin{align}
I&=\int_{\infty}^{1/6}\frac{\log(1/x)}{1+1/x^2}\,\left(\frac{-1}{x^2}\right)\,dx+\int_{1/6}^{0}\frac{\log(1/x)}{1+1/x^2}\,\left(\frac{-1}{x^2}\right)\,dx\\\\
&=\int_{\infty}^{1/6}\frac{\log(x)}{1+x^2}\,dx+\int_{1/6}^{0}\frac{\log(x)}{1+x^2}\,dx\\\\
&=-\int_0^\infty \frac{\log(x)}{1+x^2}\,dx\\\\
&=-I
\end{align}$$
Therefore, we find that $I=0$ from which we see that
$$\int_{1/6}^\infty \frac{\log(x)}{1+x^2}\,dx=-\int_0^{1/6}\frac{\log(x)}{1+x^2}\,dx$$
A: Start with the integral $I(k)$ as given by
$$I(k)=\int_0^{1/6}x^{2k}\log(x)\,dx \tag 1$$
Now, integrating by parts $(1)$ with $u=\log(x)$ and $v=\frac{x^{2k+1}}{2k+1}$ reveals
$$\begin{align}
I(k)&=-\log(6)\frac{(1/6)^{2k+1}}{2k+1}-\frac{1}{2k+1}\int_0^{1/6} x^{2k}\,dx\\\\
&=-\log(6)\frac{(1/)6^{2k+1}}{2k+1}-\frac{(1/6)^{2k+1}}{(2k+1)^2}\\\\
&=-\frac{\left(\log(6)(2k+1)+1\right)}{6^{2k+1}(2k+1)^2}
\end{align}$$
A: For the first part, note that
$$\begin{align}
-\int_{1/6}^\infty{\log x\over1+x^2}dx=\int_0^{1/6}{\log x\over1+x^2}dx
&\iff0=\int_0^{1/6}{\log x\over1+x^2}dx+\int_{1/6}^\infty{\log x\over1+x^2}dx\\
&\iff0=\int_0^\infty{\log x\over1+x^2}dx\\
&\iff0=\int_0^1{\log x\over1+x^2}dx+\int_1^\infty{\log x\over1+x^2}dx\\
&\iff-\int_1^\infty{\log x\over1+x^2}dx=\int_0^1{\log x\over1+x^2}dx
\end{align}$$
The substitution $u=1/x$ establishes the last line.
