What is the integral of 1/(2^x+7) ?

The integral of 1/(2^x+7) is x/7-1/(7ln(2))ln|2^x+7|+C

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integral of 1/(2^x+7)

\int \frac{1}{2 ^{x} + 7} d x

\frac{x}{7} - \frac{1}{7 ln ( 2 )} ln ∣ 2^{x} + 7 ∣ + C

Solution steps

\int \frac{1}{2 ^{x} + 7} d x

Apply u-substitution

: \int \frac{1}{ln ( 2 ) u ( u + 7 )} d u

= \int \frac{1}{ln ( 2 ) u ( u + 7 )} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{u ( u + 7 )} d u

Take the partial fraction of

\frac{1}{u ( u + 7 )} : \frac{1}{7 u} - \frac{1}{7 ( u + 7 )}

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{7 u} - \frac{1}{7 ( u + 7 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{ln ( 2 )} (\int \frac{1}{7 u} d u - \int \frac{1}{7 ( u + 7 )} d u)

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{1}{7 ( u + 7 )} d u = \frac{1}{7} ln ∣ u + 7 ∣

= \frac{1}{ln ( 2 )} (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = 2^{x}

= \frac{1}{ln ( 2 )} (\frac{1}{7} ln ∣ 2^{x} ∣ - \frac{1}{7} ln ∣ 2^{x} + 7 ∣)

Simplify

\frac{1}{ln ( 2 )} (\frac{1}{7} ln ∣ 2^{x} ∣ - \frac{1}{7} ln ∣ 2^{x} + 7 ∣) : \frac{x}{7} - \frac{1}{7 ln ( 2 )} ln ∣ 2^{x} + 7 ∣

= \frac{x}{7} - \frac{1}{7 ln ( 2 )} ln ∣ 2^{x} + 7 ∣

Add a constant to the solution

= \frac{x}{7} - \frac{1}{7 ln ( 2 )} ln ∣ 2^{x} + 7 ∣ + C

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What is the integral of 1/(2^x+7) ?
The integral of 1/(2^x+7) is x/7-1/(7ln(2))ln|2^x+7|+C