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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{1}{3 u - 2} 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 - 2} d u

Apply u-substitution

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

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 2 v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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