What is the integral of (e^{bt})/((a-e^{bt))^5} ?

The integral of (e^{bt})/((a-e^{bt))^5} is 1/(4b(a-e^{bt))^4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{bt})/((a-e^{bt))^5}

\int \frac{e ^{b t}}{( a - e ^{b t} ) ^{5}} d t

\frac{1}{4 b ( a - e ^{b t} ) ^{4}} + C

Solution steps

\int \frac{e ^{b t}}{( a - e ^{b t} ) ^{5}} d t

Apply u-substitution

: \int - \frac{1}{b u ^{5}} d u

= \int - \frac{1}{b u ^{5}} d u

Take the constant out:

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

= - \frac{1}{b} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= - \frac{1}{b} (- \frac{1}{4 u ^{4}})

Substitute back

u = a - e^{b t}

= - \frac{1}{b} (- \frac{1}{4 ( a - e ^{b t} ) ^{4}})

Simplify

- \frac{1}{b} (- \frac{1}{4 ( a - e ^{b t} ) ^{4}}) : \frac{1}{4 b ( a - e ^{b t} ) ^{4}}

= \frac{1}{4 b ( a - e ^{b t} ) ^{4}}

Add a constant to the solution

= \frac{1}{4 b ( a - e ^{b t} ) ^{4}} + C

\int \frac{5}{( x - 6 ) ( x - 1 )} d x

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

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

\int \frac{x ^{2} + 1}{x ^{3} + 3 x} ln (x^{3} + 3 x) d x

\int 1 n (x^{2} + 2 x + 2) d x

What is the integral of (e^{bt})/((a-e^{bt))^5} ?
The integral of (e^{bt})/((a-e^{bt))^5} is 1/(4b(a-e^{bt))^4}+C