This is the original problem.
$$\int_2^1 \frac{x^2e^x - 2xe^x}{x^4}$$
My attempt at breaking it down $$\frac{x^2e^x}{x^4} - \frac{2xe^x}{x^4}$$ $$x^{-2}e^x - 2x^{-3}e^x$$ $$ \int_2^1 x^{-2}e^x\,dx - \int_2^1 2x^{-3}e^x\,dx$$ I want to focus on the first term. $$\int_2^1 x^{-2}e^x\,dx$$
How do I do this?
The differential of $x^{-2}$ is $-2x^{-3}$
The comments on your post already solve the problem, as the integrand looks like something resulting from the quotient rule. But you can also use integration by parts for a more methodical approach. \begin{align} \int \frac{x^2e^x - 2xe^x}{x^4} dx = \int x^{-2}e^xdx - 2\int x^{-3}e^x dx \end{align} Now use integration by parts for the first part \begin{align} \int x^{-2}e^x dx = e^x x^{-2} - \int(-2x^{-3})e^x dx \end{align} Plugging this into the first equation yields \begin{align} \int \frac{x^2e^x - 2xe^x}{x^4} dx &= \int x^{-2}e^xdx - 2\int x^{-3}e^x dx\\ &= e^x x^{-2} - \int(-2x^{-3})e^x dx - 2\int x^{-3}e^x dx\\ &= \frac{e^x}{x^2} + 2\int x^{-3}e^xdx - 2\int x^{-3} e^xdx\\ &= \frac{e^x}{x^2} \end{align} This solves the indefinite integral which you can use to evaluate the definite integral.