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

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

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integral of 2xsqrt(2-x^2)

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{u}{2} d u

= 2 \cdot \int - \frac{u}{2} d u

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 2 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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

y = 2 e^{- x} + e^{3 x}

n = 1 \sum \infty \frac{n}{4 n ^{3} + 1}

h (x) = sec^{6} (x)

\int \frac{e ^{x}}{1 - 25 e ^{2 x}} d x

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