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

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

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

\int (e^{- x})^{2} d x

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

Solution steps

\int (e^{- x})^{2} d x

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = e^{- x}

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{s ^{2} + 2 s + 3}{( s + 1 ) ^{3}}

\int \frac{( 8 x ^{2} + 9 x + 8 )}{( x ^{2} + 1 ) ^{2}} d x

\int x \cdot cos (5 x) d x

x \to 0 lim (\frac{1}{1 + x ^{2}})

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