What is the integral of 1/((e^x)^2) ?

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

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integral of 1/((e^x)^2)

integral

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

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

Solution steps

\int \frac{1}{( e ^{x} ) ^{2}} d x

(e^{x})^{2} = e^{2 x}

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int e^{- 2 x} d x

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Add a constant to the solution

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

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

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

\frac{( 3 x + 10 )}{( x - 2 )}

x \cdot ln (x - 1)

x^{3} (e^{x})^{4}

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