What is the integral of (-2^{3-2x}) ?

The integral of (-2^{3-2x}) is (2^{2-2x})/(ln(2))+C

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integral of (-2^{3-2x})

\int (- 2^{3 - 2 x}) d x

\frac{2 ^{2 - 2 x}}{ln ( 2 )} + C

Solution steps

\int - 2^{3 - 2 x} d x

Take the constant out:

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

= - \int 2^{3 - 2 x} d x

2^{3 - 2 x} = 2^{3} \cdot 2^{- 2 x}

= - \int 2^{3} \cdot 2^{- 2 x} d x

Take the constant out:

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

= - 2^{3} \cdot \int 2^{- 2 x} d x

Simplify

= - 8 \cdot \int 2^{- 2 x} d x

Apply u-substitution

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

= - 8 \cdot \int - 2^{u - 1} d u

Take the constant out:

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

= - 8 (- \int 2^{u - 1} d u)

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= - 8 (- \int 2^{u} \cdot 2^{- 1} d u)

Take the constant out:

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

= - 8 (- 2^{- 1} \cdot \int 2^{u} d u)

Apply exponent rule:

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

= - 8 (- \frac{1}{2} \cdot \int 2^{u} d u)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - 8 (- \frac{1}{2} \cdot \frac{2 ^{u}}{ln ( 2 )})

Substitute back

u = - 2 x

= - 8 (- \frac{1}{2} \cdot \frac{2 ^{- 2 x}}{ln ( 2 )})

Simplify

- 8 (- \frac{1}{2} \cdot \frac{2 ^{- 2 x}}{ln ( 2 )}) : \frac{2 ^{2 - 2 x}}{ln ( 2 )}

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

Add a constant to the solution

= \frac{2 ^{2 - 2 x}}{ln ( 2 )} + C

\int \frac{y}{x} d x

\frac{\partial}{\partial y} (cos (x) + sin (y))

f (x) = 3 6 x + 8

\frac{d}{d x} (tan^{2} (sin (x)))

f (x) = x + 7

What is the integral of (-2^{3-2x}) ?
The integral of (-2^{3-2x}) is (2^{2-2x})/(ln(2))+C