What is the integral of x/(sqrt(4-x^2)) ?

The integral of x/(sqrt(4-x^2)) is -sqrt(4-x^2)+C

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integral of x/(sqrt(4-x^2))

\int \frac{x}{4 - x ^{2}} d x

- 4 - x^{2} + C

Solution steps

\int \frac{x}{4 - 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 = 4 - x^{2}

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

Simplify

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

= - 4 - x^{2}

Add a constant to the solution

= - 4 - x^{2} + C

\int 2 x + 9 d x

2 t y^{^{'}} - 3 y = 0

f (x) = \frac{7}{x ^{4}}, at x = - 3

y = x + 2, (- 1, 1)

\frac{d y}{d x} = \frac{y ^{2} + x y + x ^{2}}{x ^{2}}, y (1) = - 1

What is the integral of x/(sqrt(4-x^2)) ?
The integral of x/(sqrt(4-x^2)) is -sqrt(4-x^2)+C