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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

= \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)

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

Substitute back

u = x^{2}

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

Add a constant to the solution

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

sin ((π x)^{7})

\frac{\partial}{\partial x} (9 x + 2 y)

\int (x - 1)^{2} (x + 1)^{2} d x

\int x 1 + 3 x^{2} d x

\frac{4}{1 - x}

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