What is the integral of (3(x+4))/(x^5) ?

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

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integral of (3(x+4))/(x^5)

\int \frac{3 ( x + 4 )}{x ^{5}} d x

- \frac{1}{x ^{3}} - \frac{3}{x ^{4}} + C

Solution steps

\int \frac{3 ( x + 4 )}{x ^{5}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x + 4}{x ^{5}} d x

Expand

\frac{x + 4}{x ^{5}} : \frac{1}{x ^{4}} + \frac{4}{x ^{5}}

= 3 \cdot \int \frac{1}{x ^{4}} + \frac{4}{x ^{5}} d x

Apply the Sum Rule:

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

= 3 (\int \frac{1}{x ^{4}} d x + \int \frac{4}{x ^{5}} d x)

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

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

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

Simplify

3 (- \frac{1}{3 x ^{3}} - \frac{1}{x ^{4}}) : - \frac{1}{x ^{3}} - \frac{3}{x ^{4}}

= - \frac{1}{x ^{3}} - \frac{3}{x ^{4}}

Add a constant to the solution

= - \frac{1}{x ^{3}} - \frac{3}{x ^{4}} + C

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