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

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

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integral of (2y)/(sqrt(1-4y^2))

\int \frac{2 y}{1 - 4 y ^{2}} d y

- \frac{1}{2} 1 - 4 y^{2} + C

Solution steps

\int \frac{2 y}{1 - 4 y ^{2}} d y

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{8 u} d u

= 2 \cdot \int - \frac{1}{8 u} d u

Take the constant out:

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

= 2 (- \frac{1}{8} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 2 (- \frac{1}{8} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - 4 y^{2}

= 2 (- \frac{1}{8} \cdot 2 (1 - 4 y^{2})^{\frac{1}{2}})

Simplify

2 (- \frac{1}{8} \cdot 2 (1 - 4 y^{2})^{\frac{1}{2}}) : - \frac{1}{2} 1 - 4 y^{2}

= - \frac{1}{2} 1 - 4 y^{2}

Add a constant to the solution

= - \frac{1}{2} 1 - 4 y^{2} + C

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