What is the integral of (16x)/(sqrt(1-x^4)) ?

The integral of (16x)/(sqrt(1-x^4)) is 8arcsin(x^2)+C

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integral of (16x)/(sqrt(1-x^4))

\int \frac{16 x}{1 - x ^{4}} d x

8 arcsin (x^{2}) + C

Solution steps

\int \frac{16 x}{1 - x ^{4}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x}{1 - x ^{4}} d x

Apply u-substitution

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

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{2} arcsin (u)

Substitute back

u = x^{2}

= 16 \cdot \frac{1}{2} arcsin (x^{2})

Simplify

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

= 8 arcsin (x^{2})

Add a constant to the solution

= 8 arcsin (x^{2}) + C

\frac{d}{d x} (x^{3} + 5)

\frac{d y}{d x} = 8 x e^{7 y}

\int \frac{- 7}{x ^{2} + 16} d x

x \to 5 lim (\frac{( ln ( \frac{x}{5} ) )}{x ^{2} - 25})

\int \frac{1}{ln ( x ^{x} )} d x

What is the integral of (16x)/(sqrt(1-x^4)) ?
The integral of (16x)/(sqrt(1-x^4)) is 8arcsin(x^2)+C