The integration of ? $\int (\frac{x^{2} - 1}{x^{3} - 3 x})$ $int((x^2-1)/(x^3-3x))$ dx

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$\int (\frac{x^{2} - 1}{x^{3} - 3 x})$ $int((x^2-1)/(x^3-3x))$ dx

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Answer 1 · 2018-04-13T14:38:28

The answer is $= \frac{1}{3} ln (∣ ∣ x^{3} - 3 x ∣ ∣) + C$ $=1/3ln(|x^3-3x|)+C$

Perform this integral by substitution

Let $u = x^{3} - 3 x$ $u=x^3-3x$

$d u = 3 x^{2} - 3 = 3 (x^{2} - 1) d x$ $du=3x^2-3=3(x^2-1)dx$

Therefore, the integral is

$\int \frac{(x^{2} - 1) d x}{x^{3} - 3 x} = \frac{1}{3} \int (d u) \cdot \frac{1}{u}$ $int((x^2-1)dx)/(x^3-3x)=1/3int(du)*1/u$

$= \frac{1}{3} ln u$ $=1/3lnu$

$= \frac{1}{3} ln (∣ ∣ x^{3} - 3 x ∣ ∣) + C$ $=1/3ln(|x^3-3x|)+C$

Answer 2 · 2018-04-13T14:42:08

$\frac{1}{3} ln ∣ ∣ (x^{3} - 3 x) ∣ ∣ + C, or, ln ∣ ∣ \sqrt[3]{x^{3} - 3 x} ∣ ∣ + C$ $1/3ln|(x^3-3x)|+C, or, ln|root(3)(x^3-3x)|+C$ .

Suppose that, $I = \int \frac{x^{2} - 1}{x^{3} - 3 x} d x$ $I=int(x^2-1)/(x^3-3x)dx$ .

Observe that, $\frac{d}{d x} (x^{3} - 3 x) = 3 x^{2} - 3 = 3 (x^{2} - 1)$ $d/dx(x^3-3x)=3x^2-3=3(x^2-1)$ .

Knowing that, $int(f'(x))/f(x)dx=ln|f(x)|+c$ , we have,

$I=1/3int(3(x^2-1))/(x^3-3x)dx$ .

$rArr I=1/3ln|(x^3-3x)|+C, or, ln|root(3)(x^3-3x)|+C$ .