What is the integral of (3x^2-8x-9)/(x^3-4x^2-9x+36) ?

The integral of (3x^2-8x-9)/(x^3-4x^2-9x+36) is ln|x^3-4x^2-9x+36|+C

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integral of (3x^2-8x-9)/(x^3-4x^2-9x+36)

\int \frac{3 x ^{2} - 8 x - 9}{x ^{3} - 4 x ^{2} - 9 x + 36} d x

ln x^{3} - 4 x^{2} - 9 x + 36 + C

Solution steps

\int \frac{3 x ^{2} - 8 x - 9}{x ^{3} - 4 x ^{2} - 9 x + 36} 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} - 4 x^{2} - 9 x + 36

= ln x^{3} - 4 x^{2} - 9 x + 36

Add a constant to the solution

= ln x^{3} - 4 x^{2} - 9 x + 36 + C

x \to \infty lim (1 5^{x})

\int (3 x^{2} - 5 x)^{3} (6 x - 5) d x

y (9 - y) - \frac{81}{4} = \frac{d y}{d t}, y (0) = 4

x \to \infty lim (1. 2^{x})

x \to 0 lim (\frac{2}{1 + e ^{\frac{1}{x}}})

What is the integral of (3x^2-8x-9)/(x^3-4x^2-9x+36) ?
The integral of (3x^2-8x-9)/(x^3-4x^2-9x+36) is ln|x^3-4x^2-9x+36|+C