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

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

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

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

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

Solution steps

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

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int \frac{1}{C} d C

\frac{\partial}{\partial x} (cos^{3} (x^{7} y^{7}))

\frac{d}{d x} (0.4 x^{2})

f (x) = \frac{x ^{2}}{2 ^{x}}

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

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