# Find this definite integral <msubsup> &#x222B;<!-- ∫ --> <mrow class="MJX-TeXAtom-ORD">

Find this definite integral
${\int }_{1}^{2}\frac{{e}^{1/x}}{{x}^{4}}dx$
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Kaya Kemp
I think you made an algebra mistake when integrating by parts. Starting with
$\int {u}^{2}{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u,$
your choice of $g={u}^{2},\mathrm{d}f={e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u$ is good. This yeilds
$\int {u}^{2}{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u={u}^{2}{e}^{u}-\int 2u{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u={u}^{2}{e}^{u}-2\int u{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u.$
Next, we focus on $\int u{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u$. Like you suggested, we pick $g=u,\mathrm{d}f={e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u$. This gives us
$\int u{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u=u{e}^{u}-\int {e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u=u{e}^{u}-{e}^{u}+c.$
Putting everything together, we have
$\int {u}^{2}{e}^{u}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}u={u}^{2}{e}^{u}-2\left(u{e}^{u}-{e}^{u}+c\right)={e}^{u}\left({u}^{2}-2u+2\right)+c.$
I think you forgot to correctly distribute the factor of 2, and a minus sign.