What is the integral of 4^{2-3x} ?

The integral of 4^{2-3x} is -(2^{-6x+3})/(3ln(2))+C

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integral of 4^{2-3x}

\int 4^{2 - 3 x} d x

- \frac{2 ^{- 6 x + 3}}{3 ln ( 2 )} + C

Solution steps

\int 4^{2 - 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} \cdot 4^{u} d u

= \int - \frac{1}{3} \cdot 4^{u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int 4^{u} d u

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

= - \frac{1}{3} \cdot \frac{4 ^{u}}{ln ( 4 )}

Substitute back

u = 2 - 3 x

= - \frac{1}{3} \cdot \frac{4 ^{2 - 3 x}}{ln ( 4 )}

Simplify

- \frac{1}{3} \cdot \frac{4 ^{2 - 3 x}}{ln ( 4 )} : - \frac{2 ^{- 6 x + 3}}{3 ln ( 2 )}

= - \frac{2 ^{- 6 x + 3}}{3 ln ( 2 )}

Add a constant to the solution

= - \frac{2 ^{- 6 x + 3}}{3 ln ( 2 )} + C

\frac{48}{s ( s + 4 ) ^{2} ( s + 3 )}

\int sin (x) + 7 d x

\frac{d h}{d t} = 2 h

\int_{0}^{2} e^{x^{2}} d x

x \to 0 lim (x^{2} - 3 x)

What is the integral of 4^{2-3x} ?
The integral of 4^{2-3x} is -(2^{-6x+3})/(3ln(2))+C