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

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

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

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

4 arcsin (x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 8 \cdot \int \frac{1}{2 1 - u ^{2}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} arcsin (u)

Substitute back

u = x^{2}

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

Simplify

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

= 4 arcsin (x^{2})

Add a constant to the solution

= 4 arcsin (x^{2}) + C

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y^{^{'}} + \frac{y}{x} = 2 x^{4} y^{2}

\int e^{3 x - 8} d x

\frac{d}{d x} (e^{c o s (4 x)})

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