Determine value of integral:$I=\int_0^1\frac{\ln(1+x)}{x}dx$

Question

Determine value of integral:$$I=\int_0^1\frac{\ln(1+x)}{x}dx$$

I use Taylor's expansion with $x_0=0$, we have:

$$\ln(1+x)=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^i}{i}$$

Hence $$I=\int_0^1\frac{\ln(1+x)}{x}dx=\int_0^1\sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}dx=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i^2}$$

But, come here i don't know how! Please help me! I think the result is $I=\frac{\pi^2}{12}$

Answer 1

$$\zeta(2)=\sum_1^\infty\frac1{n^2}=\sum_1^\infty\frac1{(2n)^2}+\sum_0^\infty\frac1{(2n+1)^2}=\frac{\zeta(2)}4+\sum_0^\infty\frac1{(2n+1)^2}$$

The sum of odds is thus $\frac34$ of the total, which is $\frac{\pi^2}6$, namely $\frac{\pi^2}8$. Your sum is the difference of the two, $\frac{\pi^2}8-\frac{\pi^2}{24}=\frac{\pi^2}{12}$.