Show that $\int_0^1 \frac{\ln(1+x)}x\mathrm dx=-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx$ without actually evaluating both integrals

HINT:

Note that we have

$$\begin{align} \frac12\int_0^1 \frac{\log(x)}{1-x}\,dx&\overbrace{=}^{x\mapsto x^2}\int_0^1 \frac{x\log(x^2)}{1-x^2}\,dx\\\\ &=\int_0^1 \log(x)\left(\frac{1}{1-x}-\frac{1}{1+x}\right)\,dx \end{align}$$

Can you finish now?

Tags:

Integration

Calculus

Logarithms

Definite Integrals

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