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

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

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

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

- ln ∣ - 3 e^{x} + 2 ∣ + ln (3 ∣ - e^{x} + 1 ∣) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{v ( v + 1 )} d v

= - \int \frac{1}{v ( v + 1 )} d v

Take the partial fraction of

\frac{1}{v ( v + 1 )} : \frac{1}{v} - \frac{1}{v + 1}

= - \int \frac{1}{v} - \frac{1}{v + 1} d v

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}{v} d v - \int \frac{1}{v + 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= - (ln ∣ v ∣ - ln ∣ v + 1 ∣)

Substitute back

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

Simplify

- (ln ∣ 3 (1 - e^{x}) - 1 ∣ - ln ∣ 3 (1 - e^{x}) - 1 + 1 ∣) : - ln ∣ - 3 e^{x} + 2 ∣ + ln (3 ∣ - e^{x} + 1 ∣)

= - ln ∣ - 3 e^{x} + 2 ∣ + ln (3 ∣ - e^{x} + 1 ∣)

Add a constant to the solution

= - ln ∣ - 3 e^{x} + 2 ∣ + ln (3 ∣ - e^{x} + 1 ∣) + C

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