What is the integral of 6xsqrt(1-3x^2) ?

The integral of 6xsqrt(1-3x^2) is -2/3 (1-3x^2)^{3/2}+C

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integral of 6xsqrt(1-3x^2)

\int 6 x 1 - 3 x^{2} d x

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

Solution steps

\int 6 x 1 - 3 x^{2} d x

Take the constant out:

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

= 6 \cdot \int x 1 - 3 x^{2} d x

Apply u-substitution

: \int - \frac{u}{6} d u

= 6 \cdot \int - \frac{u}{6} d u

Take the constant out:

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

= 6 (- \frac{1}{6} \cdot \int u d u)

Apply the Power Rule

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

= 6 (- \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - 3 x^{2}

= 6 (- \frac{1}{6} \cdot \frac{2}{3} (1 - 3 x^{2})^{\frac{3}{2}})

Simplify

6 (- \frac{1}{6} \cdot \frac{2}{3} (1 - 3 x^{2})^{\frac{3}{2}}) : - \frac{2}{3} (1 - 3 x^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d}{d x} (- \frac{4}{x ^{4}})

\frac{\partial}{\partial x} (3 y^{2} e^{x} - 4 x)

\frac{\partial}{\partial x} (x^{2} yz)

\int \frac{x y}{x ^{2} + y ^{2} + 1} d x

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

What is the integral of 6xsqrt(1-3x^2) ?
The integral of 6xsqrt(1-3x^2) is -2/3 (1-3x^2)^{3/2}+C