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

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

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integral of 1/(e^x(e^x+1))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} ( u + 1 )} d u

= \int \frac{1}{u ^{2} ( u + 1 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u + 1 )} : - \frac{1}{u} + \frac{1}{u ^{2}} + \frac{1}{u + 1}

= \int - \frac{1}{u} + \frac{1}{u ^{2}} + \frac{1}{u + 1} d u

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{1}{u} d u + \int \frac{1}{u ^{2}} d u + \int \frac{1}{u + 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= - ln ∣ u ∣ - \frac{1}{u} + ln ∣ u + 1 ∣

Substitute back

u = e^{x}

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

ln ∣ e^{x} ∣ = x

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

Add a constant to the solution

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

((x - 2)^{2})^{^{'}}

y^{^{'}} = \frac{2 x}{cos ( y )}

\int_{0}^{1.25} 1000 x d x

y^{^{''}} + y^{^{'}} + 6 y = 0

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