What is the integral of (6000-2000x)/((x^2-6x+10)^2) ?

The integral of (6000-2000x)/((x^2-6x+10)^2) is (1000)/(x^2-6x+10)+C

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integral of (6000-2000x)/((x^2-6x+10)^2)

\int \frac{6000 - 2000 x}{( x ^{2} - 6 x + 10 ) ^{2}} d x

\frac{1000}{x ^{2} - 6 x + 10} + C

Solution steps

\int \frac{6000 - 2000 x}{( x ^{2} - 6 x + 10 ) ^{2}} d x

Apply u-substitution

= \int - \frac{1000}{u ^{2}} d u

Take the constant out:

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

= - 1000 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - 1000 (- \frac{1}{u})

Substitute back

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

= - 1000 (- \frac{1}{x ^{2} - 6 x + 10})

Simplify

- 1000 (- \frac{1}{x ^{2} - 6 x + 10}) : \frac{1000}{x ^{2} - 6 x + 10}

= \frac{1000}{x ^{2} - 6 x + 10}

Add a constant to the solution

= \frac{1000}{x ^{2} - 6 x + 10} + C

What is the integral of (6000-2000x)/((x^2-6x+10)^2) ?
The integral of (6000-2000x)/((x^2-6x+10)^2) is (1000)/(x^2-6x+10)+C