What is the integral of (e^{2/x})/(3x^2) ?

The integral of (e^{2/x})/(3x^2) is -1/6 e^{2/x}+C

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integral of (e^{2/x})/(3x^2)

\int \frac{e ^{\frac{2}{x}}}{3 x ^{2}} d x

- \frac{1}{6} e^{\frac{2}{x}} + C

Solution steps

\int \frac{e ^{\frac{2}{x}}}{3 x ^{2}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{e ^{\frac{2}{x}}}{x ^{2}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{2} d u

= \frac{1}{3} \cdot \int - \frac{e ^{u}}{2} d u

Take the constant out:

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

= \frac{1}{3} (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{3} (- \frac{1}{2} e^{u})

Substitute back

u = \frac{2}{x}

= \frac{1}{3} (- \frac{1}{2} e^{\frac{2}{x}})

Simplify

\frac{1}{3} (- \frac{1}{2} e^{\frac{2}{x}}) : - \frac{1}{6} e^{\frac{2}{x}}

= - \frac{1}{6} e^{\frac{2}{x}}

Add a constant to the solution

= - \frac{1}{6} e^{\frac{2}{x}} + C

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What is the integral of (e^{2/x})/(3x^2) ?
The integral of (e^{2/x})/(3x^2) is -1/6 e^{2/x}+C