Looks Familiar right??

$\begin{aligned} \int_0^\infty x^55^{-x} dx & = \int_0^\infty x^5 e^{-x \ln{5}} dx \quad \quad \color{#3D99F6}{\text{Let } u = x \ln{5} \quad \Rightarrow dx = \frac{du}{\ln{5}}} \\ & = \frac{1}{(\ln{5})^6} \int_0^\infty u^5 e^{-u} du \\ & = \frac{\Gamma (6)}{(\ln{5})^6} = \frac{5!}{(\ln{5})^6} = \frac{120}{(\ln{5})^6} \\ & \\ \Rightarrow \sqrt{a+b+c-10} & = \sqrt{120+5+6-10} = \sqrt{121} = \boxed{11} \end{aligned}$

$\large{I(n)=\displaystyle \int_{0}^{\infty} t^{n}a^{-t} dt}$

where $a$ is a constant

$let~w=t \ln a$

$dw=\ln a~ dt$

The integral now becomes

$\large{\frac{1}{(\ln a)^{n+1}} \displaystyle \int_{0}^{\infty} w^{n} e^{-w} dw}$

$\large{=\frac { \Gamma \left( n+1 \right) }{ { \left( \ln { a } \right) }^{ n+1 } } }$