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

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

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

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

2 arcsin (x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} arcsin (u)

Substitute back

u = x^{2}

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

Simplify

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

= 2 arcsin (x^{2})

Add a constant to the solution

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

\frac{d}{d x} (5 sin (x) + x^{2})

\frac{d}{d t} (740 (1.714)^{t})

\frac{d}{d x} (c \cdot e^{x})

\int 4 + (e^{x} - e^{- x})^{2} d x

f (x) = - \frac{8}{( 1 + 4 x ) ^{2}}

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