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

The integral of (e^{3x})/(2+e^x) is 1/2 (2+e^x)^2-4(2+e^x)+4ln|2+e^x|+C

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

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

\frac{1}{2} (2 + e^{x})^{2} - 4 (2 + e^{x}) + 4 ln ∣ 2 + e^{x} ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{3 ( v - 2 ) ^{2}}{v} d v

= \frac{1}{3} \cdot \int \frac{3 ( v - 2 ) ^{2}}{v} d v

Take the constant out:

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

= \frac{1}{3} \cdot 3 \cdot \int \frac{( v - 2 ) ^{2}}{v} d v

Expand

\frac{( v - 2 ) ^{2}}{v} : v - 4 + \frac{4}{v}

= \frac{1}{3} \cdot 3 \cdot \int v - 4 + \frac{4}{v} d v

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}{3} \cdot 3 (\int v d v - \int 4 d v + \int \frac{4}{v} d v)

\int v d v = \frac{v ^{2}}{2}

\int 4 d v = 4 v

\int \frac{4}{v} d v = 4 ln ∣ v ∣

= \frac{1}{3} \cdot 3 (\frac{v ^{2}}{2} - 4 v + 4 ln ∣ v ∣)

Substitute back

= \frac{1}{3} \cdot 3 \frac{( 2 + 3 e ^{3 x} ) ^{2}}{2} - 4 (2 + 3 e^{3 x}) + 4 ln 2 + 3 e^{3 x}

Simplify

\frac{1}{3} \cdot 3 \frac{( 2 + 3 e ^{3 x} ) ^{2}}{2} - 4 (2 + 3 e^{3 x}) + 4 ln 2 + 3 e^{3 x} : \frac{1}{2} (2 + e^{x})^{2} - 4 (2 + e^{x}) + 4 ln ∣ 2 + e^{x} ∣

= \frac{1}{2} (2 + e^{x})^{2} - 4 (2 + e^{x}) + 4 ln ∣ 2 + e^{x} ∣

Add a constant to the solution

= \frac{1}{2} (2 + e^{x})^{2} - 4 (2 + e^{x}) + 4 ln ∣ 2 + e^{x} ∣ + C

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What is the integral of (e^{3x})/(2+e^x) ?
The integral of (e^{3x})/(2+e^x) is 1/2 (2+e^x)^2-4(2+e^x)+4ln|2+e^x|+C