# Find the indefinite integral. \int \frac{e^{\frac{1}{t}}}{t^{2}}dt

Find the indefinite integral.
$\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt$
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Papilys3q
Step 1
Given: $I=\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt$
for evaluating given integral, in given integral we substitute
$\frac{1}{t}=p$...(1)
now, differentiating equation(1) with respect to t
so,

$=-\frac{1}{{t}^{2}}=\frac{dp}{dt}$
$\frac{dt}{{t}^{2}}=-dp$
Step 2
now, replacing $\frac{1}{t}$ with p, $\frac{dt}{{t}^{2}}$ with -dp in given integral and integrate it
so,

$=-{e}^{p}+c$...(2)
now, substitute $p=\frac{1}{t}$ in equation(2)
so,
$\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt=-{e}^{\frac{1}{t}}+c$
hence, given integral is equal to $-{e}^{\frac{1}{t}}+c$.

movingsupplyw1
Given:
$\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt$
$=-\int {e}^{u}du$
Now we calculate:
$\int {e}^{u}du$
Integral of exponential function:
$\int {a}^{u}du=\frac{{a}^{u}}{\mathrm{ln}\left(a\right)}$ at: a=e
$={e}^{u}$
We substitute the already calculated integrals:
$-\int {e}^{u}du$
$=-{e}^{u}$
$=-{e}^{\frac{1}{t}}$
$=-{e}^{\frac{1}{t}}+C$

nick1337

$\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt$
$\int -{e}^{u}du$
$-\int {e}^{u}du$
$-{e}^{u}$
$-{e}^{\frac{1}{t}}$
Result:
$-{e}^{\frac{1}{t}}+C$