What is the integral of ((e^{6x}))/(e^{6x)+1} ?

The integral of ((e^{6x}))/(e^{6x)+1} is 1/6 ln|e^{6x}+1|+C

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integral of ((e^{6x}))/(e^{6x)+1}

\int \frac{( e ^{6 x} )}{e ^{6 x} + 1} d x

\frac{1}{6} ln e^{6 x} + 1 + C

Solution steps

\int \frac{e ^{6 x}}{e ^{6 x} + 1} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = e^{6 x} + 1

= \frac{1}{6} ln e^{6 x} + 1

Add a constant to the solution

= \frac{1}{6} ln e^{6 x} + 1 + C

y = x, y = 2 x 64 - x^{2}

\frac{d y}{d x} = y^{2} e^{3 x}

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) \cdot cos (θ))

x = 6 y^{2}, x + y = 5

\frac{\partial}{\partial x} (\frac{v}{6 + uv w})

What is the integral of ((e^{6x}))/(e^{6x)+1} ?
The integral of ((e^{6x}))/(e^{6x)+1} is 1/6 ln|e^{6x}+1|+C