∫x−1x2exdx is equal to

$\int (\frac{x - 1}{x^{2}}) e^{x} d x$ is equal to

A
$e^{x} + \frac{1}{x} + c$
B
$\frac{e^{x}}{x} + c$
C
$\frac{e^{x}}{x^{2}} + c$
D
$e^{x} (\log x + \frac{1}{x}) + c$

Solution:

Let $I = \int (\frac{x - 1}{x^{2}}) e^{x} d x$

Then $I = \int [\frac{1}{x} - \frac{1}{x^{2}}] \cdot e^{x} dx$

$\int e^{x} (f (x) + f' (x)) d x = e^{x} f (x) + c$

= $e^{x} \frac{1}{x} + c$

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