What is the integral of (8x)/(sqrt(3-4x^2)) ?

The integral of (8x)/(sqrt(3-4x^2)) is -2sqrt(3-4x^2)+C

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integral of (8x)/(sqrt(3-4x^2))

\int \frac{8 x}{3 - 4 x ^{2}} d x

- 2 3 - 4 x^{2} + C

Solution steps

\int \frac{8 x}{3 - 4 x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{3 - 4 x ^{2}} d x

Apply u-substitution

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

= 8 \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

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

Apply the Power Rule

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

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

Substitute back

u = 3 - 4 x^{2}

= 8 (- \frac{1}{8} \cdot 2 (3 - 4 x^{2})^{\frac{1}{2}})

Simplify

8 (- \frac{1}{8} \cdot 2 (3 - 4 x^{2})^{\frac{1}{2}}) : - 2 3 - 4 x^{2}

= - 2 3 - 4 x^{2}

Add a constant to the solution

= - 2 3 - 4 x^{2} + C

\int_{1}^{4} (3 x^{2} + 1) d x

4 x - \frac{5 y x ^{2} + 1 d y}{d x} = 0, y (0) = - 10

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

\int \frac{sec ^{3} ( θ )}{tan ( θ )} d θ

f (x) = 16 cos (2 x)

What is the integral of (8x)/(sqrt(3-4x^2)) ?
The integral of (8x)/(sqrt(3-4x^2)) is -2sqrt(3-4x^2)+C