What is the integral of (15)/((x-1)(x^2+4)) ?

The integral of (15)/((x-1)(x^2+4)) is 15(1/5 ln|x-1|+1/5 (-1/2 ln|x^2+4|-1/2 arctan(x/2)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (15)/((x-1)(x^2+4))

\int \frac{15}{( x - 1 ) ( x ^{2} + 4 )} d x

15 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}))) + C

Solution steps

\int \frac{15}{( x - 1 ) ( x ^{2} + 4 )} d x

Take the constant out:

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

= 15 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} + 4 )} : \frac{1}{5 ( x - 1 )} + \frac{- x - 1}{5 ( x ^{2} + 4 )}

= 15 \cdot \int \frac{1}{5 ( x - 1 )} + \frac{- x - 1}{5 ( x ^{2} + 4 )} d x

Apply the Sum Rule:

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

= 15 (\int \frac{1}{5 ( x - 1 )} d x + \int \frac{- x - 1}{5 ( x ^{2} + 4 )} d x)

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

\int \frac{- x - 1}{5 ( x ^{2} + 4 )} d x = \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}))

= 15 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2})))

Add a constant to the solution

= 15 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{5} (- \frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}))) + C

f (x) = e^{5 x}

\frac{d y}{d x} = \frac{40}{x ^{- 5}}

k = 5 \sum \infty \frac{1}{( k - 2 ) ^{4}}

\frac{d}{d x} (- \frac{2 e ^{\frac{2}{x}}}{x ^{2}})

f (x) = 2 tan (x)

What is the integral of (15)/((x-1)(x^2+4)) ?
The integral of (15)/((x-1)(x^2+4)) is 15(1/5 ln|x-1|+1/5 (-1/2 ln|x^2+4|-1/2 arctan(x/2)))+C