What is the integral of (x^2)/(x^3-8) ?

The integral of (x^2)/(x^3-8) is 1/3 ln|x^3-8|+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^3-8)

\int \frac{x ^{2}}{x ^{3} - 8} d x

\frac{1}{3} ln x^{3} - 8 + C

Solution steps

\int \frac{x ^{2}}{x ^{3} - 8} 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} - 8

= \frac{1}{3} ln x^{3} - 8

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 8 + C

\int \frac{x ^{9}}{x ^{5} + 5} d x

\int \frac{( x + 5 ) ^{5}}{2 x} d x

\int \frac{1}{x ( x - 5 )} d x

\int (y e^{y} + 2) d y

\int \frac{e 5 x + 1}{5 x + 1} d x

What is the integral of (x^2)/(x^3-8) ?
The integral of (x^2)/(x^3-8) is 1/3 ln|x^3-8|+C