What is the integral of ((5(x^3+2))/(x^4)}{\frac{-30)/x} ?

The integral of ((5(x^3+2))/(x^4)}{\frac{-30)/x} is -1/6 (x-1/(x^2))+C

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integral of ((5(x^3+2))/(x^4)}{\frac{-30)/x}

\int \frac{\frac{5 ( x ^{3} + 2 )}{x ^{4}}}{\frac{- 30}{x}} d x

- \frac{1}{6} (x - \frac{1}{x ^{2}}) + C

Solution steps

\int \frac{\frac{5 ( x ^{3} + 2 )}{x ^{4}}}{\frac{- 30}{x}} d x

Take the constant out:

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

= \frac{5}{- 30} \cdot \int \frac{\frac{x ^{3} + 2}{x ^{4}}}{\frac{1}{x}} d x

\frac{5}{- 30} = - \frac{1}{6}

= (- \frac{1}{6}) \cdot \int \frac{\frac{x ^{3} + 2}{x ^{4}}}{\frac{1}{x}} d x

Simplify

\frac{\frac{x ^{3} + 2}{x ^{4}}}{\frac{1}{x}} : \frac{x ^{3} + 2}{x ^{3}}

= (- \frac{1}{6}) \cdot \int \frac{x ^{3} + 2}{x ^{3}} d x

Expand

\frac{x ^{3} + 2}{x ^{3}} : 1 + \frac{2}{x ^{3}}

= (- \frac{1}{6}) \cdot \int 1 + \frac{2}{x ^{3}} 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

= (- \frac{1}{6}) (\int 1 d x + \int \frac{2}{x ^{3}} d x)

\int 1 d x = x

\int \frac{2}{x ^{3}} d x = - \frac{1}{x ^{2}}

= (- \frac{1}{6}) (x - \frac{1}{x ^{2}})

Simplify

= - \frac{1}{6} (x - \frac{1}{x ^{2}})

Add a constant to the solution

= - \frac{1}{6} (x - \frac{1}{x ^{2}}) + C

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\int \frac{5 x ^{4} - 8 x ^{2} + 3 x - 2}{x ^{2}} d x

\int \frac{x - 1}{x ^{3} + 1} d x

\int \frac{1}{x \cdot ( 1 + x )} d x

What is the integral of ((5(x^3+2))/(x^4)}{\frac{-30)/x} ?
The integral of ((5(x^3+2))/(x^4)}{\frac{-30)/x} is -1/6 (x-1/(x^2))+C