What is the integral of (5x^3+2)/(x^3-5x^2+4x) ?

The integral of (5x^3+2)/(x^3-5x^2+4x) is 5x+1/2 ln|x|-7/3 ln|x-1|+161/6 ln|x-4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x^3+2)/(x^3-5x^2+4x)

\int \frac{5 x ^{3} + 2}{x ^{3} - 5 x ^{2} + 4 x} d x

5 x + \frac{1}{2} ln ∣ x ∣ - \frac{7}{3} ln ∣ x - 1 ∣ + \frac{161}{6} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{5 x ^{3} + 2}{x ^{3} - 5 x ^{2} + 4 x} d x

Take the partial fraction of

\frac{5 x ^{3} + 2}{x ^{3} - 5 x ^{2} + 4 x} : 5 + \frac{1}{2 x} - \frac{7}{3 ( x - 1 )} + \frac{161}{6 ( x - 4 )}

= \int 5 + \frac{1}{2 x} - \frac{7}{3 ( x - 1 )} + \frac{161}{6 ( x - 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

= \int 5 d x + \int \frac{1}{2 x} d x - \int \frac{7}{3 ( x - 1 )} d x + \int \frac{161}{6 ( x - 4 )} d x

\int 5 d x = 5 x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{7}{3 ( x - 1 )} d x = \frac{7}{3} ln ∣ x - 1 ∣

\int \frac{161}{6 ( x - 4 )} d x = \frac{161}{6} ln ∣ x - 4 ∣

= 5 x + \frac{1}{2} ln ∣ x ∣ - \frac{7}{3} ln ∣ x - 1 ∣ + \frac{161}{6} ln ∣ x - 4 ∣

Add a constant to the solution

= 5 x + \frac{1}{2} ln ∣ x ∣ - \frac{7}{3} ln ∣ x - 1 ∣ + \frac{161}{6} ln ∣ x - 4 ∣ + C

\int csc (\frac{x - π}{2}) cot (\frac{x - π}{2}) d x

x \to 0 lim (x^{8} cos (\frac{6}{x}))

\frac{d}{d x} (\frac{2}{x + 4})

\frac{100}{s ^{3} + 25 s}

\frac{d}{d x} (x^{- \frac{6}{7}})

What is the integral of (5x^3+2)/(x^3-5x^2+4x) ?
The integral of (5x^3+2)/(x^3-5x^2+4x) is 5x+1/2 ln|x|-7/3 ln|x-1|+161/6 ln|x-4|+C