What is the integral of 1/(2^{3x)} ?

The integral of 1/(2^{3x)} is -1/(3ln(2)*8^x)+C

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integral of 1/(2^{3x)}

\int \frac{1}{2 ^{3 x}} d x

- \frac{1}{3 ln ( 2 ) \cdot 8 ^{x}} + C

Solution steps

\int \frac{1}{2 ^{3 x}} d x

2^{3 x} = 8^{x}

= \int \frac{1}{8 ^{x}} d x

Apply u-substitution

: \int \frac{1}{3 ln ( 2 ) u ^{2}} d u

= \int \frac{1}{3 ln ( 2 ) u ^{2}} d u

Take the constant out:

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

= \frac{1}{3 ln ( 2 )} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{3 ln ( 2 )} (- \frac{1}{u})

Substitute back

u = 8^{x}

= \frac{1}{3 ln ( 2 )} (- \frac{1}{8 ^{x}})

Simplify

\frac{1}{3 ln ( 2 )} (- \frac{1}{8 ^{x}}) : - \frac{1}{3 ln ( 2 ) \cdot 8 ^{x}}

= - \frac{1}{3 ln ( 2 ) \cdot 8 ^{x}}

Add a constant to the solution

= - \frac{1}{3 ln ( 2 ) \cdot 8 ^{x}} + C

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What is the integral of 1/(2^{3x)} ?
The integral of 1/(2^{3x)} is -1/(3ln(2)*8^x)+C