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

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

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integral of e^{-3sqrt(x)}

\int e^{- 3 x} d x

\frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

Solution steps

\int e^{- 3 x} d x

Apply u-substitution

: \int e^{- 3 u} \cdot 2 u d u

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{v} v}{9} d v

= 2 \cdot \int \frac{e ^{v} v}{9} d v

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int e^{v} v d v

Apply Integration By Parts

: e^{v} v - \int e^{v} d v

= 2 \cdot \frac{1}{9} (e^{v} v - \int e^{v} d v)

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

= 2 \cdot \frac{1}{9} (e^{v} v - e^{v})

Substitute back

= 2 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x})

Simplify

2 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x}) : \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x})

= \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x})

Add a constant to the solution

= \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

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