Using the simmetry of the function $t \rightarrow 1/t $ to change limits of an integral. I was looking at this very famous answer and unfortunately I could not get through even the first step:
\begin{align}
& 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\
&= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ 
\end{align}
Could someone please walk me through this step? how are we using the simmetry of $1/t$ here?

In the integral
$$\int \limits _1 ^\infty \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \Bbb dt$$
make the change $u = \frac 1 t$. This will turn it into
$$\int \limits _1 ^0 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac 5 {u^2}} \right) \left( - \frac 1 {u^2} \right) \Bbb d u = \int \limits _0 ^1 \frac {u^{\frac 1 2}} {1 - \frac 1 {u^2}} \log \left( \frac {5 - \frac 2 u + \frac 1 {u^2}} {1 - \frac 2 u + \frac 5 {u^2}} \right) \left( \frac 1 {u^2} \right) \Bbb d u = \\
\int \limits _0 ^1 \frac {u^{\frac 1 2}} {u^2 - 1} \log \left( \frac {5u^2 - 2u + 1} {u^2 - 2u + 5} \right) \Bbb d u = - \int \limits _0 ^1 \frac {u^{\frac 1 2}} {u^2 - 1} \log \left( \frac {u^2 - 2u + 5} {5u^2 - 2u + 1} \right) \Bbb d u = \\
\int \limits _0 ^1 \frac {u^{\frac 1 2}} {1 - u^2} \log \left( \frac {5 - 2u + u^2} {1 - 2u + 5u^2} \right) \Bbb d u .$$
Now change the letter $u$ into $t$ and you get the second integral on the second row of your question.
