What is the integral of (x^2-2)/(sqrt(x^3-6x)) ?

The integral of (x^2-2)/(sqrt(x^3-6x)) is 2/3 sqrt(x^3-6x)+C

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integral of (x^2-2)/(sqrt(x^3-6x))

\int \frac{x ^{2} - 2}{x ^{3} - 6 x} d x

\frac{2}{3} x^{3} - 6 x + C

Solution steps

\int \frac{x ^{2} - 2}{x ^{3} - 6 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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{3} - 6 x

= \frac{1}{3} \cdot 2 (x^{3} - 6 x)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot 2 (x^{3} - 6 x)^{\frac{1}{2}} : \frac{2}{3} x^{3} - 6 x

= \frac{2}{3} x^{3} - 6 x

Add a constant to the solution

= \frac{2}{3} x^{3} - 6 x + C

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What is the integral of (x^2-2)/(sqrt(x^3-6x)) ?
The integral of (x^2-2)/(sqrt(x^3-6x)) is 2/3 sqrt(x^3-6x)+C