What is the integral of (x^9-1)/(x^4) ?

The integral of (x^9-1)/(x^4) is (x^6)/6+1/(3x^3)+C

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integral of (x^9-1)/(x^4)

\int \frac{x ^{9} - 1}{x ^{4}} d x

\frac{x ^{6}}{6} + \frac{1}{3 x ^{3}} + C

Solution steps

\int \frac{x ^{9} - 1}{x ^{4}} d x

Expand

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

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

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

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

= \frac{x ^{6}}{6} - (- \frac{1}{3 x ^{3}})

Simplify

= \frac{x ^{6}}{6} + \frac{1}{3 x ^{3}}

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{1}{3 x ^{3}} + C

\int \frac{1}{x ^{3} ln ( x )} d x

\int \frac{x ^{2} - 4 x + 1}{x ^{2} + 1} d x

\int x^{3} (2 x^{4} + 1)^{2} d x

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

\int (x^{2} - x^{4}) d x

What is the integral of (x^9-1)/(x^4) ?
The integral of (x^9-1)/(x^4) is (x^6)/6+1/(3x^3)+C