What is the integral of (6x)/(sqrt(1-x^4)) ?

The integral of (6x)/(sqrt(1-x^4)) is 3arcsin(x^2)+C

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integral of (6x)/(sqrt(1-x^4))

\int \frac{6 x}{1 - x ^{4}} d x

3 arcsin (x^{2}) + C

Solution steps

\int \frac{6 x}{1 - x ^{4}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{1 - x ^{4}} d x

Apply u-substitution

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

= 6 \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

= 6 \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)

= 6 \cdot \frac{1}{2} arcsin (u)

Substitute back

u = x^{2}

= 6 \cdot \frac{1}{2} arcsin (x^{2})

Simplify

6 \cdot \frac{1}{2} arcsin (x^{2}) : 3 arcsin (x^{2})

= 3 arcsin (x^{2})

Add a constant to the solution

= 3 arcsin (x^{2}) + C

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\frac{\partial}{\partial x} (x^{2} + y^{2} + 1)

\frac{d ^{2}}{d x ^{2}} (x e^{- x})

\frac{1 + s}{( s ( s ^{2} + s + 1 ) )}

What is the integral of (6x)/(sqrt(1-x^4)) ?
The integral of (6x)/(sqrt(1-x^4)) is 3arcsin(x^2)+C