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

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

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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

= \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 = 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

x \to 3 lim (x^{2})

\int e^{x} y d x

x \to - 1 lim (\frac{x ^{2} - 4}{x - 2})

\frac{d ^{2}}{d x ^{2}} (θ sin (θ))

\int \frac{1}{y ^{2} - 4 y + 4} d y

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