How is
$$\int_0^1 \ln \left(\frac{1}{1-x} \right) dx=1$$
using series expansion?
This is simple if one integrates it directly by first noting that the integrand is same as $-\ln(1-x)$, which can be seen here. The expansion is
$$\sum_{j=1}^{\infty} \frac{x^j}{j}$$
but integrating this, I don't see how it could produce $1$?
If you integrate, you get $$\int_0^1\left(\sum_{j=1}^\infty\frac{x^j}j\right)\,dx =\sum_{j=1}^\infty\int_0^1\frac{x^j}j\,dx=\sum_{j=1}^\infty\frac1{j(j+1)} =\sum_{j=1}^\infty\left(\frac1j-\frac1{j+1}\right)$$ which is the classic telescoping sum $$\left(1-\frac12\right)+\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right)+\cdots$$ etc.