What is the integral of (3e^{2x})/(sqrt(e^x+1)) ?

The integral of (3e^{2x})/(sqrt(e^x+1)) is 2(e^x+1)^{3/2}-6sqrt(e^x+1)+C

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integral of (3e^{2x})/(sqrt(e^x+1))

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

2 (e^{x} + 1)^{\frac{3}{2}} - 6 e^{x} + 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{u} d u

= 3 \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : u - \frac{1}{u}

= 3 \cdot \int u - \frac{1}{u} 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 u d u - \int \frac{1}{u} d u)

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

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

= 3 (\frac{2}{3} u^{\frac{3}{2}} - 2 u)

Substitute back

u = e^{x} + 1

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

Simplify

3 (\frac{2}{3} (e^{x} + 1)^{\frac{3}{2}} - 2 e^{x} + 1) : 2 (e^{x} + 1)^{\frac{3}{2}} - 6 e^{x} + 1

= 2 (e^{x} + 1)^{\frac{3}{2}} - 6 e^{x} + 1

Add a constant to the solution

= 2 (e^{x} + 1)^{\frac{3}{2}} - 6 e^{x} + 1 + C

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What is the integral of (3e^{2x})/(sqrt(e^x+1)) ?
The integral of (3e^{2x})/(sqrt(e^x+1)) is 2(e^x+1)^{3/2}-6sqrt(e^x+1)+C