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

The integral of (x^3)/(sqrt(1-4x^8)) is 1/8 arcsin(2x^4)+C

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integral of (x^3)/(sqrt(1-4x^8))

\int \frac{x ^{3}}{1 - 4 x ^{8}} d x

\frac{1}{8} arcsin (2 x^{4}) + C

Solution steps

\int \frac{x ^{3}}{1 - 4 x ^{8}} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \frac{1}{4} \cdot \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{4} \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 x^{4})

Simplify

\frac{1}{4} \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 x^{4}) : \frac{1}{8} arcsin (2 x^{4})

= \frac{1}{8} arcsin (2 x^{4})

Add a constant to the solution

= \frac{1}{8} arcsin (2 x^{4}) + C

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What is the integral of (x^3)/(sqrt(1-4x^8)) ?
The integral of (x^3)/(sqrt(1-4x^8)) is 1/8 arcsin(2x^4)+C