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

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

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integral of (6x^2-6x)/(1-3x^2+2x^3)

\int \frac{6 x ^{2} - 6 x}{1 - 3 x ^{2} + 2 x ^{3}} d x

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

Solution steps

\int \frac{6 x ^{2} - 6 x}{1 - 3 x ^{2} + 2 x ^{3}} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

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