What is the integral of 7^{3-4x} ?

The integral of 7^{3-4x} is -(7^{3-4x})/(4ln(7))+C

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integral of 7^{3-4x}

\int 7^{3 - 4 x} d x

- \frac{7 ^{3 - 4 x}}{4 ln ( 7 )} + C

Solution steps

\int 7^{3 - 4 x} d x

Apply u-substitution

: \int - \frac{1}{4} \cdot 7^{u} d u

= \int - \frac{1}{4} \cdot 7^{u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int 7^{u} d u

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

= - \frac{1}{4} \cdot \frac{7 ^{u}}{ln ( 7 )}

Substitute back

u = 3 - 4 x

= - \frac{1}{4} \cdot \frac{7 ^{3 - 4 x}}{ln ( 7 )}

Simplify

- \frac{1}{4} \cdot \frac{7 ^{3 - 4 x}}{ln ( 7 )} : - \frac{7 ^{3 - 4 x}}{4 ln ( 7 )}

= - \frac{7 ^{3 - 4 x}}{4 ln ( 7 )}

Add a constant to the solution

= - \frac{7 ^{3 - 4 x}}{4 ln ( 7 )} + C

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What is the integral of 7^{3-4x} ?
The integral of 7^{3-4x} is -(7^{3-4x})/(4ln(7))+C