Explained: "Find the indefinite integral. ∫e1tt2dt"

Juan Hewlett

Answered question

2021-12-19

Find the indefinite integral.

\int \frac{e^{\frac{1}{t}}}{t^{2}} d t

Papilys3q

Beginner2021-12-20Added 34 answers

Step 1
Given:

I = \int \frac{e^{\frac{1}{t}}}{t^{2}} d t

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}{d t} (\frac{1}{t}) = \frac{d}{d t} (p) (∵ \frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x^{2}})

= - \frac{1}{t^{2}} = \frac{d p}{d t}

\frac{d t}{t^{2}} = - d p

Step 2
now, replacing

\frac{1}{t}

with p,

\frac{d t}{t^{2}}

with -dp in given integral and integrate it
so,

\int \frac{e^{\frac{1}{t}}}{t^{2}} d t = - \int e^{p} d p (∵ \int e^{x} d x = 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}} d t = - e^{\frac{1}{t}} + c

hence, given integral is equal to

- e^{\frac{1}{t}} + c

movingsupplyw1

Beginner2021-12-21Added 30 answers

Given:

\int \frac{e^{\frac{1}{t}}}{t^{2}} d t

= - \int e^{u} d u

Now we calculate:

\int e^{u} d u

Integral of exponential function:

\int a^{u} d u = \frac{a^{u}}{\ln (a)}

at: a=e

= e^{u}

We substitute the already calculated integrals:

- \int e^{u} d u

= - e^{u}

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

Answer:

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

nick1337

Expert2021-12-28Added 777 answers

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

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