What is the integral of (-13300)/((5x+40)^2) ?

The integral of (-13300)/((5x+40)^2) is (532)/(x+8)+C

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integral of (-13300)/((5x+40)^2)

\int \frac{- 13300}{( 5 x + 40 ) ^{2}} d x

\frac{532}{x + 8} + C

Solution steps

\int \frac{- 13300}{( 5 x + 40 ) ^{2}} d x

Take the constant out:

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

= - 13300 \cdot \int \frac{1}{( 5 x + 40 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{5 u ^{2}} d u

= - 13300 \cdot \int \frac{1}{5 u ^{2}} d u

Take the constant out:

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

= - 13300 \cdot \frac{1}{5} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 13300 \cdot \frac{1}{5} (- \frac{1}{u})

Substitute back

u = 5 x + 40

= - 13300 \cdot \frac{1}{5} (- \frac{1}{5 x + 40})

Simplify

- 13300 \cdot \frac{1}{5} (- \frac{1}{5 x + 40}) : \frac{532}{x + 8}

= \frac{532}{x + 8}

Add a constant to the solution

= \frac{532}{x + 8} + C

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What is the integral of (-13300)/((5x+40)^2) ?
The integral of (-13300)/((5x+40)^2) is (532)/(x+8)+C