一个积分问题

首先，\varphi(0)=0,\varphi(1)=I

\displaystyle \varphi^\prime(\alpha)=\int_0^1\frac{x}{(1+x^2)(1+\alpha x)}dx\\{}\\ \because\frac{x}{(1+x^2)(1+\alpha x)}=\frac{1}{1+\alpha^2}\left(\frac{\alpha+x}{1+x^2}-\frac{\alpha}{1+\alpha x}\right)\\{}\\ \therefore\int_0^1\varphi^\prime(\alpha)dx=\frac{1}{1+\alpha^2}\left(\int_0^1\frac{\alpha}{1+x^2}dx+\int_0^1\frac{x}{1+x^2}dx-\int_0^1\frac{\alpha}{1+\alpha x}dx\right)\\{}\\ =\frac{1}{\alpha^2}\left[\alpha\arctan x\bigg|_0^1+\frac{1}{2}\ln(1+x^2)\bigg|_0^1-\ln(1+\alpha x)\bigg|_0^1\right]\\{}\\ =\frac{1}{\alpha^2}\left[\alpha\cdot\frac{\pi}{4}+\frac{1}{2}\ln 2-\ln(1+\alpha)\right]\\{}\\ \therefore\int_0^1\varphi^\prime(\alpha)d\alpha=\int_0^1\frac{1}{1+\alpha^2}\left[\frac{\pi}{4}\alpha+\frac{1}{2}\ln 2-\ln(1-\alpha)\right]d\alpha\\{}\\ =\frac{\pi}{8}\ln(1+\alpha^2)\bigg|_0^1+\frac{1}{2}\ln 2\arctan\alpha\bigg|_0^1-\varphi(1)\\{}\\ =\frac{\pi}{8}\ln 2+\frac{\pi}{8}\ln 2-\varphi(1)\\{}\\ =\frac{\pi}{4}\ln 2-\varphi(1)\\{}\\ \because\int_0^1\varphi^\prime(\alpha)d\alpha=\varphi(1)-\varphi(0)=\varphi(1)\\{}\\ \therefore I=\varphi(1)=\frac{\pi}{8}\ln 2