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

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

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

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

\frac{1}{4} arcsin (2 y^{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 1 - 4 u ^{2}} d u

= \int \frac{1}{2 1 - 4 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 - 4 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \frac{1}{2} \cdot \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{4} arcsin (2 y^{2}) + C

\frac{d}{d x} (4 x^{- 2 x})

\frac{d y}{d x} = 1 - x y

\int \frac{1}{x ( 100 - x )} d x

\frac{d}{d x} (1 + x + x^{2})

\frac{( s + 6 )}{s ^{2} + 6 s + 25}

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