What is the integral of x/(sqrt(9-x^2)) ?

The integral of x/(sqrt(9-x^2)) is -sqrt(9-x^2)+C

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integral of x/(sqrt(9-x^2))

\int \frac{x}{9 - x ^{2}} d x

- 9 - x^{2} + C

Solution steps

\int \frac{x}{9 - 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 = 9 - x^{2}

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

Simplify

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

= - 9 - x^{2}

Add a constant to the solution

= - 9 - x^{2} + C

4 e^{- 7 t} \cdot cos (9 t)

f (x) = sin (x) - \frac{1}{3} sin (3 x) + \frac{1}{5} sin (5 x)

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

sin (y)

\int \frac{1}{x ( x + 4 )} d x

What is the integral of x/(sqrt(9-x^2)) ?
The integral of x/(sqrt(9-x^2)) is -sqrt(9-x^2)+C