What is the integral of (10000-5000x)/((x^2-4x+5)^2) ?

The integral of (10000-5000x)/((x^2-4x+5)^2) is (2500)/(x^2-4x+5)+C

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integral of (10000-5000x)/((x^2-4x+5)^2)

\int \frac{10000 - 5000 x}{( x ^{2} - 4 x + 5 ) ^{2}} d x

\frac{2500}{x ^{2} - 4 x + 5} + C

Solution steps

\int \frac{10000 - 5000 x}{( x ^{2} - 4 x + 5 ) ^{2}} d x

Apply u-substitution

: \int - \frac{2500}{u ^{2}} d u

= \int - \frac{2500}{u ^{2}} d u

Take the constant out:

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

= - 2500 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 2500 (- \frac{1}{u})

Substitute back

u = x^{2} - 4 x + 5

= - 2500 (- \frac{1}{x ^{2} - 4 x + 5})

Simplify

- 2500 (- \frac{1}{x ^{2} - 4 x + 5}) : \frac{2500}{x ^{2} - 4 x + 5}

= \frac{2500}{x ^{2} - 4 x + 5}

Add a constant to the solution

= \frac{2500}{x ^{2} - 4 x + 5} + C

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What is the integral of (10000-5000x)/((x^2-4x+5)^2) ?
The integral of (10000-5000x)/((x^2-4x+5)^2) is (2500)/(x^2-4x+5)+C