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

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

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

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

\frac{3}{2} ln (x + 3 x - 3) + ln ∣ x ∣ - \frac{1}{2} ln x + 3 - \frac{1}{2} ln x - 3 + C

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{3 x ^{2}}{x ^{3} - 3 x} d x = \frac{3}{2} ln (x + 3 x - 3)

\int \frac{3}{x ^{3} - 3 x} d x = - ln ∣ x ∣ + \frac{1}{2} ln x + 3 + \frac{1}{2} ln x - 3

= \frac{3}{2} ln (x + 3 x - 3) - (- ln ∣ x ∣ + \frac{1}{2} ln x + 3 + \frac{1}{2} ln x - 3)

Simplify

= \frac{3}{2} ln (x + 3 x - 3) + ln ∣ x ∣ - \frac{1}{2} ln x + 3 - \frac{1}{2} ln x - 3

Add a constant to the solution

= \frac{3}{2} ln (x + 3 x - 3) + ln ∣ x ∣ - \frac{1}{2} ln x + 3 - \frac{1}{2} ln x - 3 + C

3 x \frac{d y}{d x} + 5 y = 10

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\int_{0}^{3} π (2 5 y)^{2} d y

y = F 4 x - x^{2}, (1, 3)

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