What is the integral of 7^{2x+8} ?

The integral of 7^{2x+8} is (7^{2x+8})/(2ln(7))+C

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integral of 7^{2x+8}

\int 7^{2 x + 8} d x

\frac{7 ^{2 x + 8}}{2 ln ( 7 )} + C

Solution steps

\int 7^{2 x + 8} d x

Apply u-substitution

: \int 7^{u} \cdot \frac{1}{2} d u

= \int 7^{u} \cdot \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 7^{u} d u

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

= \frac{1}{2} \cdot \frac{7 ^{u}}{ln ( 7 )}

Substitute back

u = 2 x + 8

= \frac{1}{2} \cdot \frac{7 ^{2 x + 8}}{ln ( 7 )}

Simplify

\frac{1}{2} \cdot \frac{7 ^{2 x + 8}}{ln ( 7 )} : \frac{7 ^{2 x + 8}}{2 ln ( 7 )}

= \frac{7 ^{2 x + 8}}{2 ln ( 7 )}

Add a constant to the solution

= \frac{7 ^{2 x + 8}}{2 ln ( 7 )} + C

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What is the integral of 7^{2x+8} ?
The integral of 7^{2x+8} is (7^{2x+8})/(2ln(7))+C