Show that $\int_{0}^{1} \frac{\log(1 - x^2)}{\sqrt{x} (\sqrt{x} + 1)} \,dx = \frac{7}{2} \ln(2) - \frac{5}{4} \zeta(2)$

Question

$$\int_{0}^{1} \frac{\log(1 - x^2)}{\sqrt{x} (\sqrt{x} + 1)} \,dx = \frac{7}{2} \ln(2) - \frac{5}{4} \zeta(2)$$

Here is my try \begin{align*} \text{Let: } & t = \sqrt{x} \Rightarrow dt = \frac{dx}{2\sqrt{x}} \Rightarrow dx = 2t\,dt \\ \Rightarrow & \Omega = 2 \int_{0}^{1} \frac{\ln(1 - t^4)}{t + 1} \,dt \end{align*}

$$= 2 \int_{0}^{1} \frac{\ln(1 - t^2)}{t + 1} \,dt + 2 \int_{0}^{1} \frac{\ln(1 + t^2)}{t + 1} \,dt = 2J + 2K$$

\begin{align*} K &= \int_{0}^{1} \frac{\ln(1 + t^2)}{t + 1} \,dt \\ \text{Let: } & K(a) = \int_{0}^{1} \frac{\ln(1 + at^2)}{t + 1} \,dt \Rightarrow K(a) = \int_{0}^{1} \frac{\partial}{\partial a} \left(\frac{\ln(1 + at^2)}{t + 1}\right) \,dt \end{align*}

$$= \int_{0}^{1} \frac{t^2}{(t + 1)(1 + at^2)} \,dt = \frac{1}{a + 1} \int_{0}^{1} \left(\frac{t}{a t^2 + 1} - \frac{1}{a t^2 + 1} + \frac{1}{t + 1}\right) \,dt$$

$$= \frac{\ln(a t^2 + 1) + 2a(\ln(t) + 1) - 2\sqrt{a}\arctan(t\sqrt{a})}{2a^2 + 2a} \Big|_{0}^{1}$$

Answer 1

Letting $y=\sqrt x$ transforms the integral without radical \begin{aligned} I & =\int_0^1 \frac{\ln \left(1-y^4\right)}{y(y+1)} 2 y d y \\ & =2 \int_0^1 \frac{\ln (1+y)+\ln (1-y)+\ln \left(1+y^2\right)}{y+1} d y \end{aligned}

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$$ \int_0^1 \frac{\ln (1+y)}{y+1} d y=\left[\frac{\ln ^2(1+y)}{2}\right]_0^1=\frac{\ln ^2 2}{2} $$

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By the post, we have $$ \int_0^1 \frac{\ln (1-y)}{y+1} d y=\frac{1}{2} \ln ^2 2-\frac{\pi^2}{12} $$

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By the post, we have $$\displaystyle \int_{0}^{1}\frac{\ln(1+y^{2})}{y+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$$

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Hence $$I =\int_0^1 \frac{\ln \left(1-y^4\right)}{y(y+1)} 2 y d y = \ln ^2 2+\ln ^2 2-\frac{\pi^2}{6} +\frac{3}{2}\ln^{2}(2)-\frac{{\pi}^{2}}{24}=\frac 72 \ln^2 2-\frac {5 }{4}\zeta(2) \\ $$