Finding $\int^1_0 \frac{\log(1+x)}{x}dx$ without series expansion

You might be interested in this: noticing that

$$\int_0^1 \frac{1}{1+xy}dy=\frac{\ln (x+1)}{x}$$

We can rewrite the integral as:

$$\int_0^1\frac{\ln (x+1)}{x}\;dx=\int_0^1\int_0^1 \frac{1}{1+xy}\;dy\;dx$$

Now read page 11 of this article (you'll have to slightly adapt the above of course).

Step I
Integrating by part we get that

$$\int^1_0 \frac{\log(1+x)}{x}dx=-\int^1_0 \frac{\log(x)}{x+1}dx$$

Step II
Letting $x=e^{-u}$, we have $$\int_0^{\infty}\frac{u}{e^u+1}du$$
Step III $$\int_0^{\infty}\frac{u^{s-1}}{e^u+1}du=\Gamma(s)\cdot\eta(s)\tag1$$ that is the product between gamma function and Dirichlet eta function

Step IV
Let $s=2$ in $(1)$ and we're done.

Chris.

Note that the value of $\int^1_0 \frac{n\ln(1+x^n)}{x}dx$ is invariant of $n$. Then \begin{align} \int^1_0 \frac{\ln(1+x)}{x}dx= -\frac67\int^1_0 {\frac{\ln\frac{(1+x^3)}{(1+x^2)(1+x)}}{x} } dx =-\frac67\int^1_0 \frac{\ln\left( 1-\frac x{1+x^2}\right)}{x}dx \\ \end{align} Let $J(t)=\int^1_0 \frac{\ln\left( 1-\frac {2x\sin t}{1+x^2}\right)}{x}dx$ and $$J’(t)=-\int^1_0 \frac{2\cos t\ dx }{(x-\sin t)^2 +\cos^2t}=-\left(\frac\pi2+t\right)$$ Thus $$\int^1_0 \frac{\ln(1+x)}{x}dx = -\frac67 J(\frac\pi6)= -\frac67\int_0^{\frac\pi6}J’(t)dt=\frac67\int_0^{\frac\pi6}\left(\frac\pi2+t\right)dt = \frac{\pi^2}{12}$$