What is the integral of (5x^2-5x)/(2x^3-3x^2+12) ?

The integral of (5x^2-5x)/(2x^3-3x^2+12) is 5/6 ln|2x^3-3x^2+12|+C

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integral of (5x^2-5x)/(2x^3-3x^2+12)

\int \frac{5 x ^{2} - 5 x}{2 x ^{3} - 3 x ^{2} + 12} d x

\frac{5}{6} ln 2 x^{3} - 3 x^{2} + 12 + C

Solution steps

\int \frac{5 x ^{2} - 5 x}{2 x ^{3} - 3 x ^{2} + 12} d x

Factor

\frac{5 x ^{2} - 5 x}{2 x ^{3} - 3 x ^{2} + 12}

= \int \frac{5 x ( x - 1 )}{2 x ^{3} - 3 x ^{2} + 12} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ( x - 1 )}{2 x ^{3} - 3 x ^{2} + 12} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 5 \cdot \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

= 5 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 2 x^{3} - 3 x^{2} + 12

= 5 \cdot \frac{1}{6} ln 2 x^{3} - 3 x^{2} + 12

Simplify

5 \cdot \frac{1}{6} ln 2 x^{3} - 3 x^{2} + 12 : \frac{5}{6} ln 2 x^{3} - 3 x^{2} + 12

= \frac{5}{6} ln 2 x^{3} - 3 x^{2} + 12

Add a constant to the solution

= \frac{5}{6} ln 2 x^{3} - 3 x^{2} + 12 + C

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What is the integral of (5x^2-5x)/(2x^3-3x^2+12) ?
The integral of (5x^2-5x)/(2x^3-3x^2+12) is 5/6 ln|2x^3-3x^2+12|+C