What is the integral of xsqrt(8-2x^2) ?

The integral of xsqrt(8-2x^2) is -1/6 (8-2x^2)^{3/2}+C

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integral of xsqrt(8-2x^2)

\int x 8 - 2 x^{2} d x

- \frac{1}{6} (8 - 2 x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 8 - 2 x^{2} d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 8 - 2 x^{2}

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

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (8 - 2 x^{2})^{\frac{3}{2}} : - \frac{1}{6} (8 - 2 x^{2})^{\frac{3}{2}}

= - \frac{1}{6} (8 - 2 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (8 - 2 x^{2})^{\frac{3}{2}} + C

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\int \frac{5}{8 x - x ^{2}} d x

What is the integral of xsqrt(8-2x^2) ?
The integral of xsqrt(8-2x^2) is -1/6 (8-2x^2)^{3/2}+C