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

The integral of 1/(2-e^{3x)} is -1/6 (ln|-e^{3x}+2|-3x)+C

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

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

- \frac{1}{6} (ln - e^{3 x} + 2 - 3 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{- v ^{2} + 1} d v

= - \frac{1}{3} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

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

= - \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

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

Simplify

- \frac{1}{3} (\frac{ln 2 - e ^{3 x} - 1 + 1}{2} - \frac{ln 2 - e ^{3 x} - 1 - 1}{2}) : - \frac{1}{6} (ln - e^{3 x} + 2 - 3 x)

= - \frac{1}{6} (ln - e^{3 x} + 2 - 3 x)

Add a constant to the solution

= - \frac{1}{6} (ln - e^{3 x} + 2 - 3 x) + C

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

n = 1 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

x \to - 4 + lim (\frac{x + 5}{x + 4})

2 x^{3} - x^{2} - 5 x, - x^{2} + 3 x, 0, 2

\frac{d y}{d x} = \frac{1}{4 - x ^{2}}, y (0) = π

What is the integral of 1/(2-e^{3x)} ?
The integral of 1/(2-e^{3x)} is -1/6 (ln|-e^{3x}+2|-3x)+C