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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{x + 6}{x} d x

y = 4 x + 9, (4, 5)

- 4 - 7 x^{2}

y^{^{''}} - 8 y^{^{'}} + 20 = 0

\frac{1}{2} sin (2 x)

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