What is the integral of (3-5x)/(x^3-6x^2+9x) ?

The integral of (3-5x)/(x^3-6x^2+9x) is 1/3 ln|x|-1/3 ln|x-3|+4/(x-3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3-5x)/(x^3-6x^2+9x)

\int \frac{3 - 5 x}{x ^{3} - 6 x ^{2} + 9 x} d x

\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣ + \frac{4}{x - 3} + C

Solution steps

\int \frac{3 - 5 x}{x ^{3} - 6 x ^{2} + 9 x} d x

Take the partial fraction of

\frac{3 - 5 x}{x ^{3} - 6 x ^{2} + 9 x} : \frac{1}{3 x} - \frac{1}{3 ( x - 3 )} - \frac{4}{( x - 3 ) ^{2}}

= \int \frac{1}{3 x} - \frac{1}{3 ( x - 3 )} - \frac{4}{( x - 3 ) ^{2}} 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}{3 x} d x - \int \frac{1}{3 ( x - 3 )} d x - \int \frac{4}{( x - 3 ) ^{2}} d x

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

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

\int \frac{4}{( x - 3 ) ^{2}} d x = - \frac{4}{x - 3}

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - (- \frac{4}{x - 3})

Simplify

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣ + \frac{4}{x - 3}

Add a constant to the solution

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣ + \frac{4}{x - 3} + C

\int x (1 - 2 x)^{5} d x

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

\int sin^{2} (b) x cos (b x) d x

\int x 2 x - 5 d x

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

What is the integral of (3-5x)/(x^3-6x^2+9x) ?
The integral of (3-5x)/(x^3-6x^2+9x) is 1/3 ln|x|-1/3 ln|x-3|+4/(x-3)+C