What is the integral of (8x)/(sqrt(1-4x^2)) ?

The integral of (8x)/(sqrt(1-4x^2)) is -2sqrt(1-4x^2)+C

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integral of (8x)/(sqrt(1-4x^2))

\int \frac{8 x}{1 - 4 x ^{2}} d x

- 2 1 - 4 x^{2} + C

Solution steps

\int \frac{8 x}{1 - 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}{1 - 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 = 1 - 4 x^{2}

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

Simplify

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

= - 2 1 - 4 x^{2}

Add a constant to the solution

= - 2 1 - 4 x^{2} + C

f (x) = e^{1 + 2 t a n (x)}, at x = \frac{π}{4}

t \cdot sin (k \cdot t)

3 x^{2} \frac{d y}{d x} = 2 x^{2} + y^{2}

x \to - 4 lim (- 5 x - 8)

\frac{d}{d x} (x^{2} (6 - x))

What is the integral of (8x)/(sqrt(1-4x^2)) ?
The integral of (8x)/(sqrt(1-4x^2)) is -2sqrt(1-4x^2)+C