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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u}{3 ( u - 1 )} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int \frac{v + 1}{v} d v

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

Expand

\frac{v + 1}{v} : 1 + \frac{1}{v}

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

Apply the Sum Rule:

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

= \frac{2}{3} (\int 1 d v + \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{2}{3} (v + ln ∣ v ∣)

Substitute back

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

Simplify

\frac{2}{3} (x^{3} - 1 + ln x^{3} - 1) : \frac{2}{3} (x^{\frac{3}{2}} - 1 + ln x^{\frac{3}{2}} - 1)

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

Add a constant to the solution

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

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