What is the integral of (-6000)/((3x+50)^2) ?

The integral of (-6000)/((3x+50)^2) is (2000)/(3x+50)+C

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integral of (-6000)/((3x+50)^2)

\int \frac{- 6000}{( 3 x + 50 ) ^{2}} d x

\frac{2000}{3 x + 50} + C

Solution steps

\int \frac{- 6000}{( 3 x + 50 ) ^{2}} d x

Take the constant out:

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

= - 6000 \cdot \int \frac{1}{( 3 x + 50 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2}} d u

= - 6000 \cdot \int \frac{1}{3 u ^{2}} d u

Take the constant out:

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

= - 6000 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 6000 \cdot \frac{1}{3} (- \frac{1}{u})

Substitute back

u = 3 x + 50

= - 6000 \cdot \frac{1}{3} (- \frac{1}{3 x + 50})

Simplify

- 6000 \cdot \frac{1}{3} (- \frac{1}{3 x + 50}) : \frac{2000}{3 x + 50}

= \frac{2000}{3 x + 50}

Add a constant to the solution

= \frac{2000}{3 x + 50} + C

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\int \frac{x ^{2}}{16 + x ^{2}} d x

What is the integral of (-6000)/((3x+50)^2) ?
The integral of (-6000)/((3x+50)^2) is (2000)/(3x+50)+C