What is the integral of (2x)/(sqrt(9-4x^2)) ?

The integral of (2x)/(sqrt(9-4x^2)) is -1/2 sqrt(9-4x^2)+C

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integral of (2x)/(sqrt(9-4x^2))

\int \frac{2 x}{9 - 4 x ^{2}} d x

- \frac{1}{2} 9 - 4 x^{2} + C

Solution steps

\int \frac{2 x}{9 - 4 x ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{9 - 4 x ^{2}} d x

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 = 9 - 4 x^{2}

= 2 (- \frac{1}{8} \cdot 2 (9 - 4 x^{2})^{\frac{1}{2}})

Simplify

2 (- \frac{1}{8} \cdot 2 (9 - 4 x^{2})^{\frac{1}{2}}) : - \frac{1}{2} 9 - 4 x^{2}

= - \frac{1}{2} 9 - 4 x^{2}

Add a constant to the solution

= - \frac{1}{2} 9 - 4 x^{2} + C

\frac{d}{d x} (2 - 2 sin (x))

\int_{1}^{4} (4 x^{2}) d x

x^{2} y^{^{''}} - 5 x y^{^{'}} + 9 y = 0, y (1) = - 5, y^{^{'}} (1) = - 17

(- 3.1) (0.3)

4 arcsin (x)

What is the integral of (2x)/(sqrt(9-4x^2)) ?
The integral of (2x)/(sqrt(9-4x^2)) is -1/2 sqrt(9-4x^2)+C