What is the integral of 1/(e^{2x)-3e^x} ?

The integral of 1/(e^{2x)-3e^x} is -1/9 x+1/(3e^x)+1/9 ln|e^x-3|+C

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integral of 1/(e^{2x)-3e^x}

\int \frac{1}{e ^{2 x} - 3 e ^{x}} d x

- \frac{1}{9} x + \frac{1}{3 e ^{x}} + \frac{1}{9} ln ∣ e^{x} - 3 ∣ + C

Solution steps

\int \frac{1}{e ^{2 x} - 3 e ^{x}} d x

Factor

\frac{1}{e ^{2 x} - 3 e ^{x}}

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

Apply u-substitution

: \int \frac{1}{u ^{2} ( u - 3 )} d u

= \int \frac{1}{u ^{2} ( u - 3 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u - 3 )} : - \frac{1}{9 u} - \frac{1}{3 u ^{2}} + \frac{1}{9 ( u - 3 )}

= \int - \frac{1}{9 u} - \frac{1}{3 u ^{2}} + \frac{1}{9 ( u - 3 )} 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}{9 u} d u - \int \frac{1}{3 u ^{2}} d u + \int \frac{1}{9 ( u - 3 )} d u

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{3 u ^{2}} d u = - \frac{1}{3 u}

\int \frac{1}{9 ( u - 3 )} d u = \frac{1}{9} ln ∣ u - 3 ∣

= - \frac{1}{9} ln ∣ u ∣ - (- \frac{1}{3 u}) + \frac{1}{9} ln ∣ u - 3 ∣

Substitute back

u = e^{x}

= - \frac{1}{9} ln ∣ e^{x} ∣ - (- \frac{1}{3 e ^{x}}) + \frac{1}{9} ln ∣ e^{x} - 3 ∣

Simplify

- \frac{1}{9} ln ∣ e^{x} ∣ - (- \frac{1}{3 e ^{x}}) + \frac{1}{9} ln ∣ e^{x} - 3 ∣ : - \frac{1}{9} x + \frac{1}{3 e ^{x}} + \frac{1}{9} ln ∣ e^{x} - 3 ∣

= - \frac{1}{9} x + \frac{1}{3 e ^{x}} + \frac{1}{9} ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= - \frac{1}{9} x + \frac{1}{3 e ^{x}} + \frac{1}{9} ln ∣ e^{x} - 3 ∣ + C

\frac{d}{d t} (tan (3 t))

\frac{\partial}{\partial x} (\frac{3 x ^{2}}{5 - x})

\int \frac{1}{( 1 + 9 x ^{2} )} d x

9 \cdot t + 1

\frac{d}{d x} (x e^{- \frac{x}{4}})

What is the integral of 1/(e^{2x)-3e^x} ?
The integral of 1/(e^{2x)-3e^x} is -1/9 x+1/(3e^x)+1/9 ln|e^x-3|+C