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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{5}{n + 2}

\int (x^{3} + 3) d x

e^{x} \frac{d y}{d x} = e^{- y} + e^{2 x - y}

cos (x), sin (x)

2 x^{2} y^{^{''}} + 3 x y^{^{'}} = y

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