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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 ( u - 3 u )} d u

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

Apply u-substitution

: \int \frac{3 v}{v ^{2} - 1} d v

= \frac{1}{3} \cdot \int \frac{3 v}{v ^{2} - 1} d v

Take the constant out:

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

= \frac{1}{3} \cdot 3 \cdot \int \frac{v}{v ^{2} - 1} d v

Apply u-substitution

: \int \frac{1}{2 w} d w

= \frac{1}{3} \cdot 3 \cdot \int \frac{1}{2 w} d w

Take the constant out:

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

= \frac{1}{3} \cdot 3 \cdot \frac{1}{2} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{3} \cdot 3 \cdot \frac{1}{2} ln ∣ w ∣

Substitute back

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

Simplify

\frac{1}{3} \cdot 3 \cdot \frac{1}{2} ln (3 x^{3})^{2} - 1 : \frac{1}{2} ln x^{2} - 1

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

Add a constant to the solution

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

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