What is the integral of (-8e^{2x})/(e^{2x)+3e^x+2} ?

The integral of (-8e^{2x})/(e^{2x)+3e^x+2} is -8(-ln|e^x+1|+2ln|e^x+2|)+C

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integral of (-8e^{2x})/(e^{2x)+3e^x+2}

\int \frac{- 8 e ^{2 x}}{e ^{2 x} + 3 e ^{x} + 2} d x

- 8 (- ln ∣ e^{x} + 1 ∣ + 2 ln ∣ e^{x} + 2 ∣) + C

Solution steps

\int \frac{- 8 e ^{2 x}}{e ^{2 x} + 3 e ^{x} + 2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 3 e ^{x} + 2} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 3 u + 2} d u

= - 8 \cdot \int \frac{u}{u ^{2} + 3 u + 2} d u

Take the partial fraction of

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

= - 8 \cdot \int - \frac{1}{u + 1} + \frac{2}{u + 2} 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

= - 8 (- \int \frac{1}{u + 1} d u + \int \frac{2}{u + 2} d u)

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

\int \frac{2}{u + 2} d u = 2 ln ∣ u + 2 ∣

= - 8 (- ln ∣ u + 1 ∣ + 2 ln ∣ u + 2 ∣)

Substitute back

u = e^{x}

= - 8 (- ln ∣ e^{x} + 1 ∣ + 2 ln ∣ e^{x} + 2 ∣)

Add a constant to the solution

= - 8 (- ln ∣ e^{x} + 1 ∣ + 2 ln ∣ e^{x} + 2 ∣) + C

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What is the integral of (-8e^{2x})/(e^{2x)+3e^x+2} ?
The integral of (-8e^{2x})/(e^{2x)+3e^x+2} is -8(-ln|e^x+1|+2ln|e^x+2|)+C