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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

= ln x^{3} - x + 3

Add a constant to the solution

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

\int 4 - u^{2} d u

\int - 4 x^{2} ln (- x) d x

\int e^{2} d x

\int x e^{- 0.3 x} d x

\int cos (7 x - 9) d x

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