What is the integral of 3xsqrt(1-4x^2) ?

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

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integral of 3xsqrt(1-4x^2)

\int 3 x 1 - 4 x^{2} d x

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

Solution steps

\int 3 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

= 3 \cdot \int x 1 - 4 x^{2} d x

Apply u-substitution

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

= 3 \cdot \int - \frac{u}{8} d u

Take the constant out:

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

= 3 (- \frac{1}{8} \cdot \int u d u)

Apply the Power Rule

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

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

Substitute back

u = 1 - 4 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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