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

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

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integral of 2/(3+3e^x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 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 )}

= 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

= 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 ∣

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

Substitute back

u = 3 + 3 e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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