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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((e^t-2))

integral

\frac{1}{( e ^{t} - 2 )}

- \frac{t}{2} + \frac{1}{2} ln e^{t} - 2 + C

Solution steps

\int \frac{1}{e ^{t} - 2} d t

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Substitute back

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

Simplify

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

= - \frac{t}{2} + \frac{1}{2} ln e^{t} - 2

Add a constant to the solution

= - \frac{t}{2} + \frac{1}{2} ln e^{t} - 2 + C

sec (3 x + 2) tan (3 x + 2)

x^{7} - 2 x^{5} + 2 x^{5} + x - 1

x^{\frac{1}{8}}

(2 x - 1) e^{2 x}

\frac{4}{cos ^{2} ( x )}

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