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

The integral of x/(sqrt(8-x^2)) is -sqrt(8-x^2)+C

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integral of x/(sqrt(8-x^2))

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

- 8 - x^{2} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 8 - x^{2}

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

Simplify

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

= - 8 - x^{2}

Add a constant to the solution

= - 8 - x^{2} + C

y = 3^{x}

\frac{d y}{d x} = \frac{x y + y + x + 1}{x y + 2 x - y - 2}

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

\frac{( 2 s + 1 )}{s ^{2} + 2 s + 5}

\int (x^{2} ln (x)) d x

What is the integral of x/(sqrt(8-x^2)) ?
The integral of x/(sqrt(8-x^2)) is -sqrt(8-x^2)+C