What is the integral of (x^2)/(sqrt(1-x^6)) ?

The integral of (x^2)/(sqrt(1-x^6)) is 1/3 arcsin(x^3)+C

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integral of (x^2)/(sqrt(1-x^6))

\int \frac{x ^{2}}{1 - x ^{6}} d x

\frac{1}{3} arcsin (x^{3}) + C

Solution steps

\int \frac{x ^{2}}{1 - x ^{6}} d x

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{3} arcsin (u)

Substitute back

u = x^{3}

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

Add a constant to the solution

= \frac{1}{3} arcsin (x^{3}) + C

\int \frac{x ^{2}}{4} + 2 x + 3 d x

\frac{\partial}{\partial x} (ln (1 + (x - 1)^{4} \cdot y^{2}))

\frac{\partial}{\partial y} (3 e^{x} sin (y))

\frac{d}{d x} (\frac{1 - x}{x})

\int \frac{1}{x ^{2} - 2 x + 101} d x

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