What is the integral of x/(sqrt(6-x^2)) ?

The integral of x/(sqrt(6-x^2)) is -sqrt(6-x^2)+C

integral of x/(sqrt(6-x^2))

\int \frac{x}{6 - x ^{2}} d x

- 6 - x^{2} + C

Solution steps

\int \frac{x}{6 - x ^{2}} d x

Apply u-substitution

= \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

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

Substitute back

u = 6 - x^{2}

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

Simplify

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

= - 6 - x^{2}

Add a constant to the solution

= - 6 - x^{2} + C

\frac{\partial}{\partial x} (y e^{x^{2}})

\int \frac{x + 4}{x ^{2} + 5 x - 6} d x

\frac{d y}{d x} = 6 x y

\int_{2}^{x} (3 t^{2} - 1) d t

\int \frac{csc ^{4} ( 2 x )}{cot ^{2} ( 2 x )} d x

What is the integral of x/(sqrt(6-x^2)) ?
The integral of x/(sqrt(6-x^2)) is -sqrt(6-x^2)+C