What is the integral of (e^{4/t})/(t^2) ?

The integral of (e^{4/t})/(t^2) is -1/4 e^{4/t}+C

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integral of (e^{4/t})/(t^2)

\int \frac{e ^{\frac{4}{t}}}{t ^{2}} d t

- \frac{1}{4} e^{\frac{4}{t}} + C

Solution steps

\int \frac{e ^{\frac{4}{t}}}{t ^{2}} d t

Apply u-substitution

: \int - \frac{e ^{u}}{4} d u

= \int - \frac{e ^{u}}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{4} e^{u}

Substitute back

u = \frac{4}{t}

= - \frac{1}{4} e^{\frac{4}{t}}

Add a constant to the solution

= - \frac{1}{4} e^{\frac{4}{t}} + C

\frac{d}{d x} (\frac{8 - x e ^{x}}{x + e ^{x}})

\int + \frac{1}{- x ^{2} + 8 x - 15} d x

(\frac{x ^{2}}{e ^{2 x}})^{6}

- 7 x + \frac{3}{x ^{6}}

\frac{d y}{d x} = e^{3 x - 2 y}

What is the integral of (e^{4/t})/(t^2) ?
The integral of (e^{4/t})/(t^2) is -1/4 e^{4/t}+C