What is the integral of ((17x+3))/((4x+1)(x-1)) ?

The integral of ((17x+3))/((4x+1)(x-1)) is 1/4 ln|4x+1|+4ln|x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((17x+3))/((4x+1)(x-1))

\int \frac{( 17 x + 3 )}{( 4 x + 1 ) ( x - 1 )} d x

\frac{1}{4} ln ∣ 4 x + 1 ∣ + 4 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} : \frac{1}{4 x + 1} + \frac{4}{x - 1}

= \int \frac{1}{4 x + 1} + \frac{4}{x - 1} 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

= \int \frac{1}{4 x + 1} d x + \int \frac{4}{x - 1} d x

\int \frac{1}{4 x + 1} d x = \frac{1}{4} ln ∣ 4 x + 1 ∣

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= \frac{1}{4} ln ∣ 4 x + 1 ∣ + 4 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x + 1 ∣ + 4 ln ∣ x - 1 ∣ + C

x \to 0 lim (x^{2} ln (x))

\frac{d}{d x} (\frac{1}{5 sin ( x )} + \frac{csc ( x )}{4})

\frac{d}{d x} (x - sin^{2} (x) - 4 x y e^{x y^{2}})

x y^{^{'}} + 4 y = \frac{( 2 e ^{- 4 x} )}{x ^{3}}

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

What is the integral of ((17x+3))/((4x+1)(x-1)) ?
The integral of ((17x+3))/((4x+1)(x-1)) is 1/4 ln|4x+1|+4ln|x-1|+C