What is the integral of (x^3)/(sqrt(1-x^8)) ?

The integral of (x^3)/(sqrt(1-x^8)) is 1/4 arcsin(x^4)+C

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integral of (x^3)/(sqrt(1-x^8))

\int \frac{x ^{3}}{1 - x ^{8}} d x

\frac{1}{4} arcsin (x^{4}) + C

Solution steps

\int \frac{x ^{3}}{1 - x ^{8}} d x

Apply u-substitution

: \int \frac{1}{4 1 - u ^{2}} d u

= \int \frac{1}{4 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}{4} \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}{4} arcsin (u)

Substitute back

u = x^{4}

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

Add a constant to the solution

= \frac{1}{4} arcsin (x^{4}) + C

f (x) = 3 - 4 x

y = 4 cos (x)

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

\int_{1}^{10} x ln (x) d x

x \to \infty lim (3 x - 9 x^{2} + x)

What is the integral of (x^3)/(sqrt(1-x^8)) ?
The integral of (x^3)/(sqrt(1-x^8)) is 1/4 arcsin(x^4)+C