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

The integral of ((-e^{-2x}))/2 is 1/4 e^{-2x}+C

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

integral

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

\frac{1}{4} e^{- 2 x} + C

Solution steps

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

Take the constant out:

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

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

Simplify

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

Apply u-substitution

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

= (- \frac{1}{2}) \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

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

Use the common integral:

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

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

Substitute back

u = - 2 x

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

Simplify

(- \frac{1}{2}) (- \frac{1}{2} e^{- 2 x}) : \frac{1}{4} e^{- 2 x}

= \frac{1}{4} e^{- 2 x}

Add a constant to the solution

= \frac{1}{4} e^{- 2 x} + C

\frac{lo g _{10} ( 1 )}{x}

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

\frac{( x + 8 )}{( x + 7 )}

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

\frac{81 x ^{2}}{9 x - 2}

What is the integral of ((-e^{-2x}))/2 ?
The integral of ((-e^{-2x}))/2 is 1/4 e^{-2x}+C