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

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

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integral of 1/(2+3^x)

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{u ( 2 + u )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

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

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

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

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

Substitute back

u = 3^{x}

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

Simplify

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

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

Add a constant to the solution

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

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