Prove $\int_0^1 \frac{x^{p-1}}{1-x} \log \Big(\frac{1}{x} \Big) = \sum_{n=0}^{\infty} \frac{1}{(n+p)^2}$ [duplicate]

I'm using some exercises of Apostol's Mathematical Analysis for preparing for my Integration test. However, I was struggling with the following question:

Prove that $\int_0^1 \frac{x^{p-1}}{1-x} \log (\frac{1}{x}) = \sum_{n=0}^{\infty} \frac{1}{(n+p)^2}, p>0$.

In order to prove the equality, I used Levi's monotone convergence theorem for series and the expansion of the logarithm:

$$ \int_0^1 \frac{x^{p-1}}{1-x} \log \Big(\frac{1}{x} \Big) dx= \int_0^1 - \frac{x^{p-1}}{1-x} \log(x)dx = \int_0^1 -\frac{x^{p-1}}{1-x} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n} dx = \int_0^1 \sum_{n=1}^{\infty} \frac{(1-x)^{n-1}x^{p-1}}{n}$$

By Levi's theorem on series, it follows that the integral is exactly,

$$ \sum_{n=1}^{\infty}\int_0^1 \frac{(1-x)^{n-1}x^{p-1}}{n}$$

However, I don't know how to continue with the exercise. Any hint or correction to my approach?

$$\int_0^1 \frac{x^{p-1}}{1-x} \log (\frac{1}{x})\,dx = \sum_{n=0}^{\infty} \frac{1}{(n+p)^2}, p>0$$

$$ \begin{aligned} \int_0^1 \frac{x^{p-1}}{1-x} \log (\frac{1}{x})\,dx& =-\int_0^1 \frac{x^{p-1}}{1-x} \log (x)\,dx\\ &=-\sum_{k=0}^\infty\int_0^1 x^{p+k-1} \log (x)\,dx \qquad (\text{geometric series})\\ &=-\sum_{k=0}^\infty\frac{(-1) }{(p+k)^{2}} \\ &=\sum_{k=0}^\infty\frac{1 }{(p+k)^{2}} \end{aligned} $$

Where we used

$$\int_{0}^{1} x^{m} \ln ^{n}(x) d x=\frac{(-1)^{n} n !}{(m+1)^{n+1}}$$

Proof:

$$ \begin{aligned} I &=\int_{0}^{1} x^{m} \ln ^{n}(x) d x \\ &=\int_{0}^{\infty} e^{-m x} \ln ^{n}\left(e^{-x}\right) e^{-x} d x \quad\left(x \rightarrow e^{-x}\right) \\ &=\int_{0}^{\infty} e^{-(m+1) x}(-1)^{n} x^{n} d x \\ &=\frac{(-1)^{n}}{(m+1)^{n+1}} \int_{0}^{\infty} e^{-x} x^{n} d x \quad((m+1) x \rightarrow x) \\ &=\frac{(-1)^{n} n !}{(m+1)^{n+1}} \quad \blacksquare \end{aligned} $$