What is the integral of xsqrt(1-4x^2) ?

The integral of xsqrt(1-4x^2) is -1/12 (1-4x^2)^{3/2}+C

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integral of xsqrt(1-4x^2)

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

- \frac{1}{12} (1 - 4 x^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

= \int - \frac{u}{8} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 - 4 x^{2}

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

Simplify

- \frac{1}{8} \cdot \frac{2}{3} (1 - 4 x^{2})^{\frac{3}{2}} : - \frac{1}{12} (1 - 4 x^{2})^{\frac{3}{2}}

= - \frac{1}{12} (1 - 4 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{12} (1 - 4 x^{2})^{\frac{3}{2}} + C

y^{^{''}} + 25 y = sec^{2} (5 x)

\frac{1}{1 + x ^{2}}

\frac{\partial}{\partial x} (x^{3} - x y^{2} + 3 y^{2})

\frac{d}{d x} (\frac{1}{3} π x^{2} 100 - x^{2})

\frac{( e ^{- 4 s} )}{s + 2}

What is the integral of xsqrt(1-4x^2) ?
The integral of xsqrt(1-4x^2) is -1/12 (1-4x^2)^{3/2}+C