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

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

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integral of 3/(e^x+3)

\int \frac{3}{e ^{x} + 3} d x

x - ln ∣ e^{x} + 3 ∣ + C

Solution steps

\int \frac{3}{e ^{x} + 3} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{e ^{x} + 3} d x

Apply u-substitution

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

= 3 \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 )}

= 3 \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

= 3 (\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 ∣

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

Substitute back

u = e^{x}

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

Simplify

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

= x - ln ∣ e^{x} + 3 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 3 ∣ + C

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