Inspired by Abhishek Singh's Integral!

Using a fundamental property of integrals, we rewrite our integral as

$\int_{0}^{\infty} \frac{\lbrace x \rbrace^{\lfloor x \rfloor}}{\lceil x \rceil} \ dx = \sum_{n=0}^{\infty} \int_{[n, n+1)} \frac{\lbrace x \rbrace^{\lfloor x \rfloor}}{\lceil x \rceil} \ dx$

Since we are taking the integral over the interval $[n, n+1)$ and $\lfloor x \rfloor = \lceil x \rceil -1 = n$ in the interval, we can yet again rewrite the integral as

$\sum_{n=0}^{\infty} \int_{[n, n+1)} \frac{(x-n)^{n}}{n+1} \ dx = \sum_{n=0}^{\infty} \int_{0}^{1} \frac{x^{n}}{n+1} \ dx = \sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}}$

Upon inspection, we see that this is the well-known sum found in the Basel problem, which converges to $\dfrac{\pi^{2}}{6}$ ! So finally, we conclude that

$\int_{0}^{\infty} \frac{\lbrace x \rbrace^{\lfloor x \rfloor}}{\lceil x \rceil} \ dx = \boxed{\dfrac{\pi^{2}}{6}}$

the answer should be pi^2/12 and not pi^2/6 since we have alternating +/- sign in the resolved series.. i think they have made a mistake in answer key