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}

integral

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

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

Solution steps

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

Simplify

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

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

\frac{1}{( a ^{3} - 2 a )}

4 - ∣ x ∣

(9 x + 4)^{4}

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