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

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

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integral of 1/(2^x+3)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

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

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

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

Substitute back

u = 2^{x}

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

Simplify

\frac{1}{ln ( 2 )} (\frac{1}{3} ln ∣ 2^{x} ∣ - \frac{1}{3} ln ∣ 2^{x} + 3 ∣) : \frac{x}{3} - \frac{1}{3 ln ( 2 )} ln ∣ 2^{x} + 3 ∣

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{n ( n + 5 )}

n = 0 \sum \infty \frac{i ^{n}}{3 ^{n}}

\frac{e x}{y} = 9 x - y

\int \frac{17}{x ^{3} - 125} d x

(- 2, - 3), (6, 2)

What is the integral of 1/(2^x+3) ?
The integral of 1/(2^x+3) is x/3-1/(3ln(2))ln|2^x+3|+C