Juan Hewlett

2021-12-19

Find the indefinite integral.

$\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt$

Papilys3q

Beginner2021-12-20Added 34 answers

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,

$\frac{d}{dt}\left(\frac{1}{t}\right)=\frac{d}{dt}\left(p\right)\text{}\text{}\text{}(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{{x}^{2}})$

$=-\frac{1}{{t}^{2}}=\frac{dp}{dt}$

$\frac{dt}{{t}^{2}}=-dp$

Step 2

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

so,

$\int \frac{{e}^{\frac{1}{t}}}{{t}^{2}}dt=-\int {e}^{p}dp\text{}\text{}\text{}(\because \int {e}^{x}dx={e}^{x}+c)$

$=-{e}^{p}+c$ ...(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$-{e}^{\frac{1}{t}}+c$ .

Given:

for evaluating given integral, in given integral we substitute

now, differentiating equation(1) with respect to t

so,

Step 2

now, replacing

so,

now, substitute

so,

hence, given integral is equal to

movingsupplyw1

Beginner2021-12-21Added 30 answers

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:

$-\int {e}^{u}du$

$=-{e}^{u}$

$=-{e}^{\frac{1}{t}}$

Answer:

$=-{e}^{\frac{1}{t}}+C$

Now we calculate:

Integral of exponential function:

We substitute the already calculated integrals:

Answer:

nick1337

Expert2021-12-28Added 777 answers

Add C

Result:

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