Reasoning in an integration substitution 
Evaluate: $\displaystyle\int_{0}^{1}\frac{\ln(x+1)}{x^2+1}\,\mathrm{d}x$

So I did this a completely different way than what the answer key states. I used integration by parts and some symmetry tricks and got the correct answer. However the answer key says: 

Make the substitution $x=\frac{1-u}{1+u}$

The same solution was reached in about half the steps but still using symmetry, my questions are How would I know to do that? Is this a certain type of substitution? Is there something else that maybe I could use this for?

Here's how I would do it. Let $$f(\alpha)=\int_0^1 \frac{\log(1+\alpha x)}{1+x^2}\ dx$$ where $f(1)$ is the integral we seek to evaluate.
By differentiation under the integral sign we have $$f'(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{\log(1+\alpha x)}{1+x^2}\ dx\\ =\int_0^1\frac{x}{(1+x^2)(1+\alpha x)}\ dx.$$
Now we apply partial fraction decomposition: suppose that $$\frac{x}{(1+x^2)(1+\alpha x)}=\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}$$ for some constants $A,B,C$. By combining the fractions and collecting the terms we obtain the linear system of equations $$\begin{cases}A+\alpha B=1\\ \alpha A+C=0\\B+C=0\end{cases}$$ which we find to be equivalent to $$\begin{cases}A=\frac{1}{1+\alpha^2}\\B=\frac{\alpha}{1+\alpha^2}\\C=-\frac{\alpha}{1+\alpha^2}\end{cases}$$ From this we conclude that $$f'(\alpha)=\int_0^1\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}\ dx\\ =\left[\frac12 A\log(1+x^2)+B\arctan{x}+\frac{C}{\alpha}\log(1+\alpha x)\right]_{x=0}^1\\ =\frac{\log2}{2}\frac{1}{1+\alpha^2}+\frac{\pi}{4} \frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}$$ It is easy to see by looking at the definition that $f(0)=0$. From this it follows that $f(\alpha_0)=\int_0^{\alpha_0} f'(\alpha)\ d\alpha$, and in particular $$f(1)=\int_0^1 \frac{\log2}{\alpha}\frac{1}{1+\alpha^2}+\frac{\pi}{4}\frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}\ d\alpha$$ which simplifies to $$\color{red}{\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx}= \frac{\pi \log 2}{16}-\color{red}{\int_0^1 \frac{\log(1+\alpha)}{1+\alpha^2}\ d\alpha}$$ hence the result $$\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx=\frac{\pi \log 2}{8}.$$
