What is the integral of (x^9-3x^4)/(x^6) ?

The integral of (x^9-3x^4)/(x^6) is (x^4)/4+3/x+C

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integral of (x^9-3x^4)/(x^6)

\int \frac{x ^{9} - 3 x ^{4}}{x ^{6}} d x

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

Solution steps

\int \frac{x ^{9} - 3 x ^{4}}{x ^{6}} d x

Expand

\frac{x ^{9} - 3 x ^{4}}{x ^{6}} : x^{3} - \frac{3}{x ^{2}}

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

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

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

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

Simplify

= \frac{x ^{4}}{4} + \frac{3}{x}

Add a constant to the solution

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

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What is the integral of (x^9-3x^4)/(x^6) ?
The integral of (x^9-3x^4)/(x^6) is (x^4)/4+3/x+C