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

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

Popular >

integral of (x^2-2x)/(x^3-3x^2)

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

\frac{1}{3} ln x^{3} - 3 x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 3 x^{2}

= \frac{1}{3} ln x^{3} - 3 x^{2}

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 3 x^{2} + C

- 3 t (6 t^{3} - 1)^{8}

\int_{0}^{π} x sin (x) d x

\frac{d}{d x} ((- 2 x^{2} - 2 x + 3)^{5})

\int cos (x) sin (x) cos^{2} (x) + 1 d x

\int_{0}^{x} e^{x^{2}} x d x

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