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

The integral of (e^{-3/x})/(3x^2) is 1/9 e^{-3/x}+C

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

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

\frac{1}{9} e^{- \frac{3}{x}} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{3} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{3} e^{u}

Substitute back

u = - \frac{3}{x}

= \frac{1}{3} \cdot \frac{1}{3} e^{- \frac{3}{x}}

Simplify

\frac{1}{3} \cdot \frac{1}{3} e^{- \frac{3}{x}} : \frac{1}{9} e^{- \frac{3}{x}}

= \frac{1}{9} e^{- \frac{3}{x}}

Add a constant to the solution

= \frac{1}{9} e^{- \frac{3}{x}} + C

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What is the integral of (e^{-3/x})/(3x^2) ?
The integral of (e^{-3/x})/(3x^2) is 1/9 e^{-3/x}+C