What is the integral of (e^{9/t})/(t^2) ?

The integral of (e^{9/t})/(t^2) is -1/9 e^{9/t}+C

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integral of (e^{9/t})/(t^2)

\int \frac{e ^{\frac{9}{t}}}{t ^{2}} d t

- \frac{1}{9} e^{\frac{9}{t}} + C

Solution steps

\int \frac{e ^{\frac{9}{t}}}{t ^{2}} d t

Apply u-substitution

= \int - \frac{e ^{u}}{9} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{9} e^{u}

Substitute back

u = \frac{9}{t}

= - \frac{1}{9} e^{\frac{9}{t}}

Add a constant to the solution

= - \frac{1}{9} e^{\frac{9}{t}} + C

What is the integral of (e^{9/t})/(t^2) ?
The integral of (e^{9/t})/(t^2) is -1/9 e^{9/t}+C