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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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