Evaluate $\int_{2}^{\infty} \frac{x^2-x}{6^x} dx$

Question

Can someone please help me to solve

$$\int_{2}^{\infty} \frac{x^2-x}{6^x} \mathrm dx ?$$

I don't know how to integrate it as there is no exponential involved but the constant $6$.

Answer 1

Expanding on the comments. There is an exponential function.

Recall that for $a>0$,

$$a^x = (e^{\ln a})^x = e^{x\ln (a) },$$

so the integral can be rewritten as

$$\int_{2}^{\infty} (x^2-x)e^{-x\ln(6)} \mathrm dx.$$

Got it from there?