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

The integral of 2xsqrt(1-3x^2) is -2/9 (-3x^2+1)^{3/2}+C

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

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

- \frac{2}{9} (- 3 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int 2 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

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

Apply u-substitution

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

= 2 \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

= 2 (- \frac{1}{6} \cdot \int u d u)

Apply the Power Rule

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

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

Substitute back

u = 1 - 3 x^{2}

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

Simplify

2 (- \frac{1}{6} \cdot \frac{2}{3} (1 - 3 x^{2})^{\frac{3}{2}}) : - \frac{2}{9} (- 3 x^{2} + 1)^{\frac{3}{2}}

= - \frac{2}{9} (- 3 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{9} (- 3 x^{2} + 1)^{\frac{3}{2}} + C

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What is the integral of 2xsqrt(1-3x^2) ?
The integral of 2xsqrt(1-3x^2) is -2/9 (-3x^2+1)^{3/2}+C