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

The integral of (6x^3-6x)/(1-3x^2+2x^3) is 3x+4ln|x-1|+1/2 ln|2x+1|+C

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

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

3 x + 4 ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣ + C

Solution steps

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

Long division

\frac{6 x ^{3} - 6 x}{1 - 3 x ^{2} + 2 x ^{3}} : 3 + \frac{3 ( 3 x + 1 )}{( x - 1 ) ( 2 x + 1 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 3 d x = 3 x

\int \frac{3 ( 3 x + 1 )}{( x - 1 ) ( 2 x + 1 )} d x = 4 ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣

= 3 x + 4 ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= 3 x + 4 ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ 2 x + 1 ∣ + C

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What is the integral of (6x^3-6x)/(1-3x^2+2x^3) ?
The integral of (6x^3-6x)/(1-3x^2+2x^3) is 3x+4ln|x-1|+1/2 ln|2x+1|+C