What is the integral of ((4x^{19}-5x^{-8})/(x^2)) ?

The integral of ((4x^{19}-5x^{-8})/(x^2)) is (2x^{18})/9+5/(9x^9)+C

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integral of ((4x^{19}-5x^{-8})/(x^2))

\int (\frac{4 x ^{19} - 5 x ^{- 8}}{x ^{2}}) d x

\frac{2 x ^{18}}{9} + \frac{5}{9 x ^{9}} + C

Solution steps

\int \frac{4 x ^{19} - 5 x ^{- 8}}{x ^{2}} d x

Expand

\frac{4 x ^{19} - 5 x ^{- 8}}{x ^{2}} : 4 x^{17} - \frac{5}{x ^{10}}

= \int 4 x^{17} - \frac{5}{x ^{10}} 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

= \int 4 x^{17} d x - \int \frac{5}{x ^{10}} d x

\int 4 x^{17} d x = \frac{2 x ^{18}}{9}

\int \frac{5}{x ^{10}} d x = - \frac{5}{9 x ^{9}}

= \frac{2 x ^{18}}{9} - (- \frac{5}{9 x ^{9}})

Simplify

= \frac{2 x ^{18}}{9} + \frac{5}{9 x ^{9}}

Add a constant to the solution

= \frac{2 x ^{18}}{9} + \frac{5}{9 x ^{9}} + C

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What is the integral of ((4x^{19}-5x^{-8})/(x^2)) ?
The integral of ((4x^{19}-5x^{-8})/(x^2)) is (2x^{18})/9+5/(9x^9)+C