Problem Statement
Prove that $$\sum_{k=1}^{\infty}\frac{1}{(k+p)^2}=-\int_0^1\frac{x^p}{1-x}\log x \ \mathrm{d}x$$
Background
I just learned the limit theorems (MCT, LDCT, Fatou's Lemma). This problem is the same as this one except that I wish for a solution using the tools I know, rather than complex analysis.
Attempt First, recalling the geometric series, \begin{align*} -\int_0^1\frac{x^p}{1-x}\log x \ \mathrm{d}x&=-\int_0^1 \lim_{n\to\infty} \sum_{k=0}^{n}x^{k+p} \log x \ \mathrm{d}x \\ &=-\lim_{n \to \infty}\int_0^1\sum_{k=0}^{n}x^{k+p} \log x \ \mathrm{d}x, \end{align*} where the limits were interchanged by invoking the "reverse" Monotone Convergence after realizing that $\require{cancel}\sum_{k=0}^{n}x^{k+p} \log x \geq \sum_{k=0}^{n+1}x^{k+p} \log x.$ Next, we use integration by parts as follows: \begin{align*} -\lim_{n \to \infty}\int_0^1\sum_{k=0}^{n}x^{k+p} \log x \ \mathrm{d}x&=-\lim_{n \to \infty}\int_0^1\log x \ \mathrm{d}\left(\sum_{k=1}^{n}\frac{x^{k+p}}{k+p} \right)\\ &=-\lim_{n\to\infty}\left(\cancelto{0}{\log(1)\sum_{k=1}^{n}\frac{x^{k+p}}{k+p}}-\cancelto{0}{\lim_{x\to 0}\log(x)\sum_{k=1}^{n}\frac{x^{k+p}}{k+p}}\\ -\int_0^1\sum_{k=1}^{n}\frac{1}{x}\frac{x^{k+p}}{k+p}\ \mathrm{d}x\right)\\ &=\lim_{n\to\infty}\int_0^1\sum_{k=1}^n \frac{x^{k+p-1}}{k+p} \\ &=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{(k+p)^2}, \end{align*} as desired.
Question
Are my steps rigorous and justified enough for a first year graduate real analysis course?