What is the integral of-(2200x)/(sqrt(25-x^2)) ?

The integral of-(2200x)/(sqrt(25-x^2)) is 2200sqrt(25-x^2)+C

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integral of-(2200x)/(sqrt(25-x^2))

\int - \frac{2200 x}{25 - x ^{2}} d x

2200 25 - x^{2} + C

Solution steps

\int - \frac{2200 x}{25 - x ^{2}} d x

Take the constant out:

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

= - 2200 \cdot \int \frac{x}{25 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{2 u} d u

= - 2200 \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

= - 2200 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - 2200 (- \frac{1}{2} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 25 - x^{2}

= - 2200 (- \frac{1}{2} \cdot 2 (25 - x^{2})^{\frac{1}{2}})

Simplify

- 2200 (- \frac{1}{2} \cdot 2 (25 - x^{2})^{\frac{1}{2}}) : 2200 25 - x^{2}

= 2200 25 - x^{2}

Add a constant to the solution

= 2200 25 - x^{2} + C

x \to 0 lim (x ln (x^{3}))

\frac{x - y}{x + y}

y^{^{''}} + 9 y^{^{'}} + 14 y = 0

f (x) = ln (\frac{x + 2}{x ^{2}}), at x = 1

\frac{d}{d x} (\frac{( ln ( x ) ) ^{3}}{x ^{3}})

What is the integral of-(2200x)/(sqrt(25-x^2)) ?
The integral of-(2200x)/(sqrt(25-x^2)) is 2200sqrt(25-x^2)+C