What is the integral of (e^{1/t})/(t^2) ?

The integral of (e^{1/t})/(t^2) is -e^{1/t}+C

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integral of (e^{1/t})/(t^2)

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

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

Solution steps

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

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - e^{u}

Substitute back

u = \frac{1}{t}

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

Add a constant to the solution

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

\frac{d}{d x} (2 x^{2} - x)

\frac{2 s + 16}{s ^{2} + 4 s + 13}

- 18 x^{2} y^{^{''}} + 4 x (x + 9) y^{^{'}} - 4 (x + 9) y = - 4 x^{3}

x \to 9 lim (x)

\int \frac{x}{25 + x ^{2}} d x

What is the integral of (e^{1/t})/(t^2) ?
The integral of (e^{1/t})/(t^2) is -e^{1/t}+C