TITLE: Integral of $x^{-2}e^x$
QUESTION [2 upvotes]: 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 $-2x^{-3}$

REPLY [2 votes]: 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.
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