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

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

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integral of (x^2-2x)/(x^3-3x^2+1)

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

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

Solution steps

\int \frac{x ^{2} - 2 x}{x ^{3} - 3 x ^{2} + 1} 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} + 1

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

Add a constant to the solution

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

\frac{d}{d x} (e^{s i n (x)} \cdot cos (sin (x)))

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{18}}

\int X^{2} d X

\frac{\partial}{\partial x} (2 y^{3} + 4 x)

\int \frac{2 - 8 x}{9 - 4 x ^{2}} d x

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