What is the integral of (x^5)/(sqrt(1-x^3)) ?

The integral of (x^5)/(sqrt(1-x^3)) is -1/9 (-2(-x^3+1)^{3/2}+6sqrt(-x^3+1))+C

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integral of (x^5)/(sqrt(1-x^3))

\int \frac{x ^{5}}{1 - x ^{3}} d x

- \frac{1}{9} (- 2 (- x^{3} + 1)^{\frac{3}{2}} + 6 - x^{3} + 1) + C

Solution steps

\int \frac{x ^{5}}{1 - x ^{3}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{5}}}{5 - u ^{\frac{3}{5}} + 1} d u

= \int \frac{u ^{\frac{1}{5}}}{5 - u ^{\frac{3}{5}} + 1} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{u ^{\frac{1}{5}}}{- u ^{\frac{3}{5}} + 1} d u

Apply u-substitution

: \int - \frac{5 ( - v + 1 )}{3 v} d v

= \frac{1}{5} \cdot \int - \frac{5 ( - v + 1 )}{3 v} d v

Take the constant out:

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

= \frac{1}{5} (- \frac{5}{3} \cdot \int \frac{- v + 1}{v} d v)

Expand

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

= \frac{1}{5} (- \frac{5}{3} \cdot \int - v + \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{1}{5} (- \frac{5}{3} (- \int v d v + \int \frac{1}{v} d v))

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

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

= \frac{1}{5} (- \frac{5}{3} (- \frac{2}{3} v^{\frac{3}{2}} + 2 v))

Substitute back

= \frac{1}{5} (- \frac{5}{3} (- \frac{2}{3} (- (x^{5})^{\frac{3}{5}} + 1)^{\frac{3}{2}} + 2 - (x^{5})^{\frac{3}{5}} + 1))

Simplify

\frac{1}{5} (- \frac{5}{3} (- \frac{2}{3} (- (x^{5})^{\frac{3}{5}} + 1)^{\frac{3}{2}} + 2 - (x^{5})^{\frac{3}{5}} + 1)) : - \frac{1}{9} (- 2 (- x^{3} + 1)^{\frac{3}{2}} + 6 - x^{3} + 1)

= - \frac{1}{9} (- 2 (- x^{3} + 1)^{\frac{3}{2}} + 6 - x^{3} + 1)

Add a constant to the solution

= - \frac{1}{9} (- 2 (- x^{3} + 1)^{\frac{3}{2}} + 6 - x^{3} + 1) + C

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What is the integral of (x^5)/(sqrt(1-x^3)) ?
The integral of (x^5)/(sqrt(1-x^3)) is -1/9 (-2(-x^3+1)^{3/2}+6sqrt(-x^3+1))+C