What is the integral of 1/(8^x) ?

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

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integral of 1/(8^x)

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

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

Solution steps

\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

f (x) = 5 x^{2}, at x = 0

n = 1 \sum \infty (- 1)^{n + 1} \frac{n + 1}{2 n ^{2} + 1}

\int \frac{ln ( x ^{\frac{7}{2}} )}{x} d x

\frac{d}{d x} ((ln (x))^{p (x)})

\frac{d y}{d x} = \frac{y}{( x ^{2} + 1 )}