What is the integral of 2/(3e^{-5x)-4} ?

The integral of 2/(3e^{-5x)-4} is -1/10 (ln|3e^{-5x}-4|-ln(3)+5x)+C

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integral of 2/(3e^{-5x)-4}

\int \frac{2}{3 e ^{- 5 x} - 4} d x

- \frac{1}{10} (ln 3 e^{- 5 x} - 4 - ln (3) + 5 x) + C

Solution steps

\int \frac{2}{3 e ^{- 5 x} - 4} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{3 e ^{- 5 x} - 4} d x

Apply u-substitution

: \int - \frac{1}{5 u ( u + 4 )} d u

= 2 \cdot \int - \frac{1}{5 u ( u + 4 )} d u

Take the constant out:

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

= 2 (- \frac{1}{5} \cdot \int \frac{1}{u ( u + 4 )} d u)

Take the partial fraction of

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= 2 (- \frac{1}{5} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 4 )} 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

= 2 (- \frac{1}{5} (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u))

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

= 2 (- \frac{1}{5} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣))

Substitute back

u = 3 e^{- 5 x} - 4

= 2 (- \frac{1}{5} (\frac{1}{4} ln 3 e^{- 5 x} - 4 - \frac{1}{4} ln 3 e^{- 5 x} - 4 + 4))

Simplify

2 (- \frac{1}{5} (\frac{1}{4} ln 3 e^{- 5 x} - 4 - \frac{1}{4} ln 3 e^{- 5 x} - 4 + 4)) : - \frac{1}{10} (ln 3 e^{- 5 x} - 4 - ln (3) + 5 x)

= - \frac{1}{10} (ln 3 e^{- 5 x} - 4 - ln (3) + 5 x)

Add a constant to the solution

= - \frac{1}{10} (ln 3 e^{- 5 x} - 4 - ln (3) + 5 x) + C

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What is the integral of 2/(3e^{-5x)-4} ?
The integral of 2/(3e^{-5x)-4} is -1/10 (ln|3e^{-5x}-4|-ln(3)+5x)+C