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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + e^{3 x}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{- x ^{2} + 4}{( x ^{2} + 4 ) ^{2}})

\int 2 t^{5} d t

y^{^{'}} = - \frac{y}{2}

y^{^{''}} - 8 y^{^{'}} = - 2 x - 1

2 x^{3} - 3 x^{2} + x - 1

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