how to evaluate the integral $\int_0^1 \frac{\ln(1+x)}{x} dx$？

Question

How do I evaluate the integral:

$$ \int_0^1 \frac{\ln(1+x)}{x} dx $$

in a simple way, without using complex methods?

Answer 1

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$

$$\frac{\ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$$

$$\int_{0}^{1} \frac{\ln(1+x)}{x}\ dx=\large(x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\ldots\large) \Biggr|_{0}^{1}$$

$$= 1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+ \frac{1}{5^2}\ldots$$

$$= 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+ \frac{1}{5^2}\ldots - 2(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2} \dots) $$

$$ = \frac{\pi^2}{6}-\frac{2}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}\ldots)$$

$$ = \frac{\pi^2}{6}-\frac{1}{2}(\frac{\pi^2}{6}) = \frac{\pi^2}{12}$$