I have recently found that $\zeta(2)$ can be found by integrating $\frac{\log(1+x)}{x}$ from $-1$ to $1$. Since $$ \int_{-1}^{1}\frac{\log(1+x)}{x}dx=\int_{-1}^{1}(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)dx=2*\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\frac{3}{2}\zeta(2)$$ $\frac{\log(1+x)}{x}$ has a pole in $-1$, hence it is an improper integral. This can be avoided by taking a contour that will circle around that. This can be further be extended into other positive integers by noting this formulation $$\frac{1}{y}\int_{0}^{y}\frac{\log(1+x)}{x}dx=(1-\frac{y}{2^2}+\frac{y^2}{3^2}-\frac{y^3}{4^2}+...)$$ If we integrate it again from $-1$ to $1$ we get $$ \int_{-1}^{1}(\frac{1}{y}\int_{0}^{y}\frac{\log(1+x)}{x}dx)dy=\int_{-1}^{1}(1-\frac{y}{2^2}+\frac{y^2}{3^2}-\frac{y^3}{4^2}+...)dy=2*\sum_{n=0}^{\infty} \frac{1}{(2n+1)^3}=2(1-\frac{1}{2^3})\zeta(3)$$ Similarly, we can find values of $\zeta(k)$ for all positive integers if we can integrate $$ \int_{-1}^{1}(\frac{1}{y_k}...\int_{0}^{y_3}(\frac{1}{y_2}\int_{0}^{y_2}(\frac{1}{y_1}\int_{0}^{y_1}\frac{\log(1+x)}{x}dx)dy_1)dy_2)....)dy_k$$
Is there any general way to solve the integral mentioned above?