What is the integral of (e^{6/t})/(t^2) ?

The integral of (e^{6/t})/(t^2) is -1/6 e^{6/t}+C

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integral of (e^{6/t})/(t^2)

\int \frac{e ^{\frac{6}{t}}}{t ^{2}} d t

- \frac{1}{6} e^{\frac{6}{t}} + C

Solution steps

\int \frac{e ^{\frac{6}{t}}}{t ^{2}} d t

Apply u-substitution

: \int - \frac{e ^{u}}{6} d u

= \int - \frac{e ^{u}}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{6} e^{u}

Substitute back

u = \frac{6}{t}

= - \frac{1}{6} e^{\frac{6}{t}}

Add a constant to the solution

= - \frac{1}{6} e^{\frac{6}{t}} + C

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} - 3 x z)

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

\frac{d y}{d x}, 2 x^{2} + y^{2} - 7 = 0

\int \frac{1}{x ^{2} + 2 x + 26} d x

What is the integral of (e^{6/t})/(t^2) ?
The integral of (e^{6/t})/(t^2) is -1/6 e^{6/t}+C