What is the integral of (20x)/(sqrt(1-x^4)) ?

The integral of (20x)/(sqrt(1-x^4)) is 10arcsin(x^2)+C

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integral of (20x)/(sqrt(1-x^4))

\int \frac{20 x}{1 - x ^{4}} d x

10 arcsin (x^{2}) + C

Solution steps

\int \frac{20 x}{1 - x ^{4}} d x

Take the constant out:

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

= 20 \cdot \int \frac{x}{1 - x ^{4}} d x

Apply u-substitution

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

= 20 \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

= 20 \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)

= 20 \cdot \frac{1}{2} arcsin (u)

Substitute back

u = x^{2}

= 20 \cdot \frac{1}{2} arcsin (x^{2})

Simplify

20 \cdot \frac{1}{2} arcsin (x^{2}) : 10 arcsin (x^{2})

= 10 arcsin (x^{2})

Add a constant to the solution

= 10 arcsin (x^{2}) + C

\int \frac{1}{4 + 2 x} d x

y = x^{3} + 7 x - 5, (1, 3)

\int \frac{8 x ^{3}}{( x ^{2} + 2 ) ^{2}} d x

\int (\frac{8}{x}) d x

\frac{d y}{d x} = \frac{1}{x ^{8}}

What is the integral of (20x)/(sqrt(1-x^4)) ?
The integral of (20x)/(sqrt(1-x^4)) is 10arcsin(x^2)+C