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

= \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

y = (x^{2} + 1)

\int \frac{36 + ( ln ( 8 x ) ) ^{2}}{x} d x

x \to 0 lim (\frac{x ^{2} - 9}{x + 3})

y^{^{'}} = \frac{( x + y + 1 )}{( 2 x + 2 y + 3 )}

\frac{d}{d x} (\frac{4 x - 3}{5 x ^{2} + 3 x})

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