What is the integral of (e^x)/(e^{3x)+e^{2x}} ?

The integral of (e^x)/(e^{3x)+e^{2x}} is -x-1/(e^x)+ln|e^x+1|+C

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integral of (e^x)/(e^{3x)+e^{2x}}

\int \frac{e ^{x}}{e ^{3 x} + e ^{2 x}} d x

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

Solution steps

\int \frac{e ^{x}}{e ^{3 x} + e ^{2 x}} d x

Simplify

\frac{e ^{x}}{e ^{3 x} + e ^{2 x}} : \frac{1}{e ^{x} ( e ^{x} + 1 )}

= \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

\frac{\partial}{\partial y} (∣ y ∣ + \frac{1}{2 a} ∣ x - y ∣^{2})

\int \frac{1}{4 x + 1} d x

\frac{\partial}{\partial x} (\frac{7}{x})

x \to 4 + lim (\frac{x + 1}{x - 4})

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 2)

What is the integral of (e^x)/(e^{3x)+e^{2x}} ?
The integral of (e^x)/(e^{3x)+e^{2x}} is -x-1/(e^x)+ln|e^x+1|+C