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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{v ( v - 2 )} d v

= - \int \frac{1}{v ( v - 2 )} d v

Complete the square

v (v - 2) : (v - 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{w ^{2} - 1} d w

= - \int \frac{1}{w ^{2} - 1} d w

Factor

w^{2} - 1 : - (- w^{2} + 1)

= - \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

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

= - (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= - (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

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