I've tried to find this integral by the method already outlined in the title. I decided to let $$ \displaystyle I(\alpha) = \int_{0}^{1} \dfrac{\log(1+\alpha x)}{1+x^2} \text{ d}x. $$ From this integral, I differentiated the equality with respect to $\alpha$ and performed a partial fraction decomposition on the resulting integral.
$$\begin{aligned} I'(\alpha) \ \ & = \int_{0}^{1} \dfrac{\partial}{\partial \alpha} \left( \dfrac{\log(1+\alpha x)}{1+x^2} \right) \text{ d}x \\ & = \int_{0}^{1} \dfrac{x}{(1+\alpha x)(1+x^2)} \text{ d}x \\ & = \dfrac{-\alpha}{1+\alpha^2} \int_{0}^{1} \dfrac{1}{1+\alpha x} \text{ d}x + \dfrac{1}{1+\alpha^2} \int_{0}^{1} \dfrac{x+\alpha}{1+x^2} \text{ d}x \\ & = \dfrac{-1}{1+\alpha^2} \log \left| 1+\alpha x\right| \Big|_{\ x=0}^{\ x=1} + \dfrac{1}{2(1+\alpha^2)} \log \left( 1+x^2 \right) \Big|_{\ x=0}^{\ x=1} + \dfrac{\alpha}{1+\alpha^2} \arctan x \Big|_{\ x=0}^{\ x=1} \\ & = \dfrac{-\log \left| 1 + \alpha \right|}{1+\alpha^2} + \dfrac{\log(2)}{2(1+\alpha^2)} + \dfrac{\alpha \pi}{4(1+\alpha^2)} \end{aligned} $$
Integrating this equality with respect to $\alpha$, I get something along the lines of.
$$\begin{aligned} I(\alpha) \ \ & = - \int \dfrac{-\log \left| 1 + \alpha \right|}{1+\alpha^2} \text{ d}\alpha + \dfrac{\log(2)}{2} \arctan \alpha + \dfrac{\pi}{8} \log(1+\alpha^2) + \mathcal{C} \end{aligned}$$
Where $\mathcal{C}$ is a constant of integration to be found.
Now from here on in, I'm unsure of how to perform integration by parts on the integral in the final line, with $u=\log|1+\alpha|$ so that I can finish the integral off.
Any help would be greatly appreciated!
Noticing from the definition of $I(\alpha)$ that $I(0)=0$, from your last line we obtain $$I(\alpha)=-\int_0^\alpha \frac{\log(1+\alpha)}{1+\alpha^2}\,d\alpha+\frac{\log2}{2}\arctan\alpha+\frac{\pi}{8}\log(1+\alpha^2),\quad \alpha>0.$$ What we need is $I(1)$, so let's set in this formula $\alpha=1$: $$I(1)=-I(1)+\frac{\pi\log2}{4}.$$ Hence, $I(1)=\dfrac{\pi\log2}{8}$.