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

The integral of 5/(3x^3-9x^2) is 5(1/27 (-ln|x|+3/x)+1/27 ln|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 5/(3x^3-9x^2)

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

5 (\frac{1}{27} (- ln ∣ x ∣ + \frac{3}{x}) + \frac{1}{27} ln ∣ x - 3 ∣) + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{1}{3 x ^{3} - 9 x ^{2}} d x

Take the partial fraction of

\frac{1}{3 x ^{3} - 9 x ^{2}} : \frac{- x - 3}{27 x ^{2}} + \frac{1}{27 ( x - 3 )}

= 5 \cdot \int \frac{- x - 3}{27 x ^{2}} + \frac{1}{27 ( x - 3 )} 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

= 5 (\int \frac{- x - 3}{27 x ^{2}} d x + \int \frac{1}{27 ( x - 3 )} d x)

\int \frac{- x - 3}{27 x ^{2}} d x = \frac{1}{27} (- ln ∣ x ∣ + \frac{3}{x})

\int \frac{1}{27 ( x - 3 )} d x = \frac{1}{27} ln ∣ x - 3 ∣

= 5 (\frac{1}{27} (- ln ∣ x ∣ + \frac{3}{x}) + \frac{1}{27} ln ∣ x - 3 ∣)

Add a constant to the solution

= 5 (\frac{1}{27} (- ln ∣ x ∣ + \frac{3}{x}) + \frac{1}{27} ln ∣ x - 3 ∣) + C

\int \frac{4 x}{2 x + 6} d x

\int ln (x) d ln (x) d x

\int 4 (x^{2} + x + 2) e^{- 2 x} d x

\int (csc^{2} (y) + 1) d y

\int e^{x} sin (a x) d x

What is the integral of 5/(3x^3-9x^2) ?
The integral of 5/(3x^3-9x^2) is 5(1/27 (-ln|x|+3/x)+1/27 ln|x-3|)+C