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

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

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

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

\frac{2}{3} arcsin (x^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{1 - u ^{6}} d u

= \int \frac{2 u ^{2}}{1 - u ^{6}} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{1 - u ^{6}} d u

Apply u-substitution

: \int \frac{1}{3 1 - v ^{2}} d v

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 - v ^{2}} d v = arcsin (v)

= 2 \cdot \frac{1}{3} arcsin (v)

Substitute back

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

Simplify

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

= \frac{2}{3} arcsin (x^{\frac{3}{2}})

Add a constant to the solution

= \frac{2}{3} arcsin (x^{\frac{3}{2}}) + C

\int (3 x + 1) e^{2 x} d x

n = 1 \sum \infty n (n - 1) x^{n}

\frac{d y}{d x} = y - x - 1 + (x - y + 2)^{- 1}

\int \frac{ln ( x )}{9} d x

\frac{\partial}{\partial x} (x z + yz)

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