<s>[INST] Evaluate the improper integral $ \int_0^1 \frac{\ln(1+x)}{x}\,dx $ I am trying to evaluate 
$$
\int_0^1 \frac{\ln(1+x)}{x}\,dx
$$
I started by using the Taylor series for $\ln (1+x)$
$$\begin{align*}
\int_0^1 \frac{\ln(1+x)}{x}\,dx &= \int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}dx \\&=\int_0^1 \frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+...\right)dx \\&=\int_0^1 \left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}-\frac{x^5}{6}+...\right)dx\\&=1-\frac{1}{4}+\frac{1}{9} -\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+... \\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\end{align*}
$$
I'm aware of the fact that $$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$$ however the $(-1)^{n+1}$ is giving me trouble.
 [/INST] Hint: If you take $\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}$ what do you get?
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