What is the integral of (3x^2-2x-9)/(x^3-x^2-9x+9) ?

The integral of (3x^2-2x-9)/(x^3-x^2-9x+9) is ln|x^3-x^2-9x+9|+C

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integral of (3x^2-2x-9)/(x^3-x^2-9x+9)

\int \frac{3 x ^{2} - 2 x - 9}{x ^{3} - x ^{2} - 9 x + 9} d x

ln x^{3} - x^{2} - 9 x + 9 + C

Solution steps

\int \frac{3 x ^{2} - 2 x - 9}{x ^{3} - x ^{2} - 9 x + 9} 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 = x^{3} - x^{2} - 9 x + 9

= ln x^{3} - x^{2} - 9 x + 9

Add a constant to the solution

= ln x^{3} - x^{2} - 9 x + 9 + C

\int \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}} d x

(- 8.8) (- 4.3)

\int e^{5 x} sin (x) d x

x y^{^{'}} = y + 2 x^{2} sin (x)

\frac{d}{d x} (g (x))

What is the integral of (3x^2-2x-9)/(x^3-x^2-9x+9) ?
The integral of (3x^2-2x-9)/(x^3-x^2-9x+9) is ln|x^3-x^2-9x+9|+C