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

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

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

\frac{1}{3} ln x^{3} - 2 + C

Solution steps

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

= \frac{1}{3} ln x^{3} - 2

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 2 + C

x = - 2, x = 3, y = x^{2} - 12, y = x - 6

1 - x^{12}

\frac{\partial ^{2}}{\partial x ^{2}} (\frac{1}{x y})

\int x sin (\frac{1}{6} x) d x

\int \frac{4}{x - 4} d x

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