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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{3} ln x^{3} - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} - 1

= \frac{1}{3} ln x^{3} - 1

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 1 + C

\frac{d}{d x} (ln (x + 7))

\frac{d}{d x} (3^{t a n (x)})

\frac{d y}{d x} = \frac{y ^{2} + 2 x y}{x ^{2}}

\int \frac{1 - 2 x}{x ( 1 - x )} d x

\frac{- 4 x}{x ^{2} + 1}

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