What is the integral of (4x^2)/(sqrt(1-x^6)) ?

The integral of (4x^2)/(sqrt(1-x^6)) is 4/3 arcsin(x^3)+C

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integral of (4x^2)/(sqrt(1-x^6))

\int \frac{4 x ^{2}}{1 - x ^{6}} d x

\frac{4}{3} arcsin (x^{3}) + C

Solution steps

\int \frac{4 x ^{2}}{1 - x ^{6}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{1 - x ^{6}} d x

Apply u-substitution

= 4 \cdot \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

= 4 \cdot \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)

= 4 \cdot \frac{1}{3} arcsin (u)

Substitute back

u = x^{3}

= 4 \cdot \frac{1}{3} arcsin (x^{3})

Simplify

4 \cdot \frac{1}{3} arcsin (x^{3}) : \frac{4}{3} arcsin (x^{3})

= \frac{4}{3} arcsin (x^{3})

Add a constant to the solution

= \frac{4}{3} arcsin (x^{3}) + C

What is the integral of (4x^2)/(sqrt(1-x^6)) ?
The integral of (4x^2)/(sqrt(1-x^6)) is 4/3 arcsin(x^3)+C