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

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

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

integral

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

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

Solution steps

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

Take the constant out:

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

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

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

= e \cdot \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}

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

Apply u-substitution

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

= e \cdot \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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{x}{( 4 + 2 x )}

5^{2 x - 7}

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

\frac{1}{3 x ^{2} - 3 x}

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