What is the integral of (-1800x)/(sqrt(25-x^2)) ?

The integral of (-1800x)/(sqrt(25-x^2)) is 1800sqrt(25-x^2)+C

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integral of (-1800x)/(sqrt(25-x^2))

\int \frac{- 1800 x}{25 - x ^{2}} d x

1800 25 - x^{2} + C

Solution steps

\int \frac{- 1800 x}{25 - x ^{2}} d x

Take the constant out:

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

= - 1800 \cdot \int \frac{x}{25 - x ^{2}} d x

Apply u-substitution

= - 1800 \cdot \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

= - 1800 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

= - 1800 (- \frac{1}{2} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 25 - x^{2}

= - 1800 (- \frac{1}{2} \cdot 2 (25 - x^{2})^{\frac{1}{2}})

Simplify

- 1800 (- \frac{1}{2} \cdot 2 (25 - x^{2})^{\frac{1}{2}}) : 1800 25 - x^{2}

= 1800 25 - x^{2}

Add a constant to the solution

= 1800 25 - x^{2} + C

l a pl a ce t r an s f or m t^{2} cos (3 t)

x \to 13 lim (x^{4} - 10 x^{2} - 20)

x \to \infty lim (x e^{(\frac{1}{x})})

in v erse l a pl a ce \frac{s ^{2} + 4}{( s ^{2} - 4 ) ^{2}}

5 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

What is the integral of (-1800x)/(sqrt(25-x^2)) ?
The integral of (-1800x)/(sqrt(25-x^2)) is 1800sqrt(25-x^2)+C