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

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

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

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

- \frac{1}{6} ln 1 + 3 x^{2} - 2 x^{3} + C

Solution steps

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

Apply u-substitution

= \int - \frac{1}{6 u} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{6} ln ∣ u ∣

Substitute back

u = 1 + 3 x^{2} - 2 x^{3}

= - \frac{1}{6} ln 1 + 3 x^{2} - 2 x^{3}

Add a constant to the solution

= - \frac{1}{6} ln 1 + 3 x^{2} - 2 x^{3} + C

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