What is the integral of (-1400x)/(sqrt(25-x^2)) ?

The integral of (-1400x)/(sqrt(25-x^2)) is 1400sqrt(25-x^2)+C

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integral of (-1400x)/(sqrt(25-x^2))

\int \frac{- 1400 x}{25 - x ^{2}} d x

1400 25 - x^{2} + C

Solution steps

\int \frac{- 1400 x}{25 - x ^{2}} d x

Take the constant out:

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

= - 1400 \cdot \int \frac{x}{25 - x ^{2}} d x

Apply u-substitution

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

= - 1400 \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

= - 1400 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= - 1400 (- \frac{1}{2} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 25 - x^{2}

= - 1400 (- \frac{1}{2} \cdot 2 (25 - x^{2})^{\frac{1}{2}})

Simplify

- 1400 (- \frac{1}{2} \cdot 2 (25 - x^{2})^{\frac{1}{2}}) : 1400 25 - x^{2}

= 1400 25 - x^{2}

Add a constant to the solution

= 1400 25 - x^{2} + C

\frac{d}{d x} (sin (x) cos (x) + sin (x))

x^{3} + x^{2}

x \to \infty lim (\frac{m}{n})

n = 0 \sum \infty 0.0 1^{n}

y + y^{^{'}} = sin (t), y (π) = 1

What is the integral of (-1400x)/(sqrt(25-x^2)) ?
The integral of (-1400x)/(sqrt(25-x^2)) is 1400sqrt(25-x^2)+C