What is the integral of (4e^{3x})/(1+e^{3x)} ?

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

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integral of (4e^{3x})/(1+e^{3x)}

\int \frac{4 e ^{3 x}}{1 + e ^{3 x}} d x

\frac{4}{3} ln 1 + e^{3 x} + C

Solution steps

\int \frac{4 e ^{3 x}}{1 + e ^{3 x}} d x

Take the constant out:

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

= 4 \cdot \int \frac{e ^{3 x}}{1 + e ^{3 x}} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= 4 \cdot \int \frac{1}{3 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + e^{3 x}

= 4 \cdot \frac{1}{3} ln 1 + e^{3 x}

Simplify

4 \cdot \frac{1}{3} ln 1 + e^{3 x} : \frac{4}{3} ln 1 + e^{3 x}

= \frac{4}{3} ln 1 + e^{3 x}

Add a constant to the solution

= \frac{4}{3} ln 1 + e^{3 x} + C

x \to 0 lim (1 + 3^{x})

\int x^{2} x^{3} + 7 d x

\frac{d y}{d x}, 2 x^{2} + 2 x + x y = 1

f (x) = e^{x} - x^{e}

\frac{\partial}{\partial x} (A^{2} x^{2} + B^{2} y^{2})

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