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

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

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

integral

(e^{- x})^{3}

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

Solution steps

\int (e^{- x})^{3} d x

Simplify

(e^{- x})^{3} : e^{- 3 x}

= \int e^{- 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - 3 x

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

Add a constant to the solution

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

sin^{6} (x)

(e^{t})^{2}

e^{\frac{1}{x}} x

\frac{( x - 1 ) ^{2}}{x ^{2}}

\frac{t}{6}

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