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

The integral of (x^5)/(sqrt(1-3x^3)) is 2/27 (1/3 (1-3x^3)^{3/2}-sqrt(1-3x^3))+C

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

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

\frac{2}{27} (\frac{1}{3} (1 - 3 x^{3})^{\frac{3}{2}} - 1 - 3 x^{3}) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 ( u ^{2} - 1 )}{27} d u

= \int \frac{2 ( u ^{2} - 1 )}{27} d u

Take the constant out:

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

= \frac{2}{27} \cdot \int u^{2} - 1 d u

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}{27} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= \frac{2}{27} (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 - 3 x^{3}

= \frac{2}{27} (\frac{( 1 - 3 x ^{3} ) ^{3}}{3} - 1 - 3 x^{3})

Simplify

\frac{2}{27} (\frac{( 1 - 3 x ^{3} ) ^{3}}{3} - 1 - 3 x^{3}) : \frac{2}{27} (\frac{1}{3} (1 - 3 x^{3})^{\frac{3}{2}} - 1 - 3 x^{3})

= \frac{2}{27} (\frac{1}{3} (1 - 3 x^{3})^{\frac{3}{2}} - 1 - 3 x^{3})

Add a constant to the solution

= \frac{2}{27} (\frac{1}{3} (1 - 3 x^{3})^{\frac{3}{2}} - 1 - 3 x^{3}) + C

\frac{\partial}{\partial x} (t e^{- 2 x} + cos (x))

4 - \frac{3}{x ^{2}}, at x = 4

f (x) = \frac{2 x - 1}{x + 9}, at x = 1

f (x) = \frac{( 4 - 2 x ) ^{5}}{x ^{3} + 12}

What is the integral of (x^5)/(sqrt(1-3x^3)) ?
The integral of (x^5)/(sqrt(1-3x^3)) is 2/27 (1/3 (1-3x^3)^{3/2}-sqrt(1-3x^3))+C