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

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

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

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

\frac{x ^{6}}{6} - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x ^{9} - x ^{3}}{x ^{4}} : x^{5} - \frac{1}{x}

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

\int x^{5} d x = \frac{x ^{6}}{6}

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{6}}{6} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{6}}{6} - ln ∣ x ∣ + C

\int \frac{1}{x} - \frac{2}{9 x ^{2} + 1} d x

\int \frac{( 9 x ^{2} )}{1 - x ^{3}} d x

\int \frac{1}{( x + 1 ) ^{2} + 2} d x

\int \frac{1}{( u ) ^{2}} d u

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

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