What is the integral of (5(-3r-4))/(r^5) ?

The integral of (5(-3r-4))/(r^5) is 5(1/(r^3)+1/(r^4))+C

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integral of (5(-3r-4))/(r^5)

\int \frac{5 ( - 3 r - 4 )}{r ^{5}} d r

5 (\frac{1}{r ^{3}} + \frac{1}{r ^{4}}) + C

Solution steps

\int \frac{5 ( - 3 r - 4 )}{r ^{5}} d r

Take the constant out:

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

= 5 \cdot \int \frac{- 3 r - 4}{r ^{5}} d r

Expand

\frac{- 3 r - 4}{r ^{5}} : - \frac{3}{r ^{4}} - \frac{4}{r ^{5}}

= 5 \cdot \int - \frac{3}{r ^{4}} - \frac{4}{r ^{5}} d r

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (- \int \frac{3}{r ^{4}} d r - \int \frac{4}{r ^{5}} d r)

\int \frac{3}{r ^{4}} d r = - \frac{1}{r ^{3}}

\int \frac{4}{r ^{5}} d r = - \frac{1}{r ^{4}}

= 5 (- (- \frac{1}{r ^{3}}) - (- \frac{1}{r ^{4}}))

Simplify

= 5 (\frac{1}{r ^{3}} + \frac{1}{r ^{4}})

Add a constant to the solution

= 5 (\frac{1}{r ^{3}} + \frac{1}{r ^{4}}) + C

What is the integral of (5(-3r-4))/(r^5) ?
The integral of (5(-3r-4))/(r^5) is 5(1/(r^3)+1/(r^4))+C