What is the integral of (9000-3000x)/((x^2-6x+10)^2) ?

The integral of (9000-3000x)/((x^2-6x+10)^2) is (1500)/(x^2-6x+10)+C

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integral of (9000-3000x)/((x^2-6x+10)^2)

\int \frac{9000 - 3000 x}{( x ^{2} - 6 x + 10 ) ^{2}} d x

\frac{1500}{x ^{2} - 6 x + 10} + C

Solution steps

\int \frac{9000 - 3000 x}{( x ^{2} - 6 x + 10 ) ^{2}} d x

Apply u-substitution

: \int - \frac{1500}{u ^{2}} d u

= \int - \frac{1500}{u ^{2}} d u

Take the constant out:

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

= - 1500 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 1500 (- \frac{1}{u})

Substitute back

u = x^{2} - 6 x + 10

= - 1500 (- \frac{1}{x ^{2} - 6 x + 10})

Simplify

- 1500 (- \frac{1}{x ^{2} - 6 x + 10}) : \frac{1500}{x ^{2} - 6 x + 10}

= \frac{1500}{x ^{2} - 6 x + 10}

Add a constant to the solution

= \frac{1500}{x ^{2} - 6 x + 10} + C

\frac{d}{d x} ((2 x + 3)^{3})

f (x) = \frac{x}{x + 3}, at x = - 5

9 x^{2} \cdot cos (x) \cdot cot (x)

\frac{\partial}{\partial x} (r (θ, x) sin (θ))

x = 5, y = 5 x

What is the integral of (9000-3000x)/((x^2-6x+10)^2) ?
The integral of (9000-3000x)/((x^2-6x+10)^2) is (1500)/(x^2-6x+10)+C