What is the integral of 9/(3e^{-2v)-7} ?

The integral of 9/(3e^{-2v)-7} is -9/14 (ln|3e^{-2v}-7|-ln(3)+2v)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/(3e^{-2v)-7}

\int \frac{9}{3 e ^{- 2 v} - 7} d v

- \frac{9}{14} (ln 3 e^{- 2 v} - 7 - ln (3) + 2 v) + C

Solution steps

\int \frac{9}{3 e ^{- 2 v} - 7} d v

Take the constant out:

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

= 9 \cdot \int \frac{1}{3 e ^{- 2 v} - 7} d v

Apply u-substitution

: \int - \frac{1}{2 u ( u + 7 )} d u

= 9 \cdot \int - \frac{1}{2 u ( u + 7 )} d u

Take the constant out:

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

= 9 (- \frac{1}{2} \cdot \int \frac{1}{u ( u + 7 )} d u)

Take the partial fraction of

\frac{1}{u ( u + 7 )} : \frac{1}{7 u} - \frac{1}{7 ( u + 7 )}

= 9 (- \frac{1}{2} \cdot \int \frac{1}{7 u} - \frac{1}{7 ( u + 7 )} d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (- \frac{1}{2} (\int \frac{1}{7 u} d u - \int \frac{1}{7 ( u + 7 )} d u))

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{1}{7 ( u + 7 )} d u = \frac{1}{7} ln ∣ u + 7 ∣

= 9 (- \frac{1}{2} (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣))

Substitute back

u = 3 e^{- 2 v} - 7

= 9 (- \frac{1}{2} (\frac{1}{7} ln 3 e^{- 2 v} - 7 - \frac{1}{7} ln 3 e^{- 2 v} - 7 + 7))

Simplify

9 (- \frac{1}{2} (\frac{1}{7} ln 3 e^{- 2 v} - 7 - \frac{1}{7} ln 3 e^{- 2 v} - 7 + 7)) : - \frac{9}{14} (ln 3 e^{- 2 v} - 7 - ln (3) + 2 v)

= - \frac{9}{14} (ln 3 e^{- 2 v} - 7 - ln (3) + 2 v)

Add a constant to the solution

= - \frac{9}{14} (ln 3 e^{- 2 v} - 7 - ln (3) + 2 v) + C

\int \frac{e ^{2 x}}{1 + 6 e ^{2 x} + 9 e ^{4 x}} d x

\int sin (7 x) cos (8 x) d x

\int \frac{ln ( 3 x )}{x ^{11}} d x

\int 5 sin (3 x) - 4 x sec (x^{2}) d x

\int \frac{x ^{2} + 1}{x ^{3} + 4 x ^{2} + 3 x} d x

What is the integral of 9/(3e^{-2v)-7} ?
The integral of 9/(3e^{-2v)-7} is -9/14 (ln|3e^{-2v}-7|-ln(3)+2v)+C