Evaluate $\int_0^1(x\ln(x))^{50}dx$

Hint. Following your approach by differentiation under the integral sign, note that $$\int_{0}^{1} x^t (\ln(x))^n \, dx=\frac{d^n}{dt^n}\left( \int_0^1 x^t dx\right)=\frac{d^n}{dt^n} \left((t+1)^{-1}\right).$$

I thought it might be instructive to present an approach that is efficient, elementary, and circumvents differentiation under the integral. To that end, we proceed.

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The integral can be evaluated straightforwardly by enforcing the substitution $x\mapsto e^{-x/(n+1)}$. Using this substitution, we have for any integer $n\ge 0$

$$\begin{align} \int_0^1 x^n\log^n(x)\,dx&=\frac{(-1)^n}{(n+1)^{n+1}}\underbrace{\int_0^\infty x^ne^{-x}\,dx}_{n!}\\\\ &=\frac{(-1)^nn!}{(n+1)^{n+1}} \end{align}$$

Then, let $n=50$.

Hint:

By parts,

$$\int_0^1x^{50}\log^{50}xdx=\left.\frac{x^{51}}{51}\log^{50}x\right|_0^1-\frac{50}{51}\int_0^1x^{50}\log x^{49}dx.$$

The first term evaluates to $0$, and the second shows an easy recurrence.

After $50$ iterations,

$$\frac{50!}{51^{50}}\int_0^1 x^{50}dx.$$