Parametric improper integral

Let $\alpha, \beta>0$ such that $\alpha \neq \beta$ . Prove that

$\int_{0}^{\infty} \frac{1-x^2}{\left ( \alpha x +\beta \right )^2} \frac{\ln\left ( 1+x \right )}{\left ( \beta x + \alpha \right )^2} \, \mathrm{d}x = \frac{1}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \ln \frac{\beta}{\alpha}$

Solution

Let $\mathcal{J}$ denote the integral. Thus,

$\begin{align*} \mathcal{J}&=\int_0^\infty \frac{(1-x^2)\ln(1+x)}{(\alpha x+\beta)^2(\alpha+\betax)^2} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{x\to\frac{1}{x}}{=\! =\! =\! =\! =\!} -\int_0^\infty \frac{(1-x^2)(\ln(1+x)-\ln x)}{(\alpha x+\beta)^2(\alpha+\beta x)^2} \, \mathrm{d}x \\ &=\frac{1}{2} \int_0^\infty \frac{(1-x^2)\ln x}{(\alpha x+\beta)^2(\alpha+\beta x)^2} \, \mathrm{d} x\\ &=\frac{1}{2\left (\alpha^2-\beta^2 \right )}\int_0^\infty \frac{\ln x}{(\alpha x+\beta)^2} \, \mathrm{d} x- \\ &\quad \quad - \frac{1}{2\left (\alpha^2-\beta^2 \right )}\int_0^\infty \frac{\ln x}{(\alpha+\beta x)^2} \, \mathrm{d}x \end{align*}$

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