What is the integral of (1/(x^9)-x^9-1/2) ?

The integral of (1/(x^9)-x^9-1/2) is -1/(8x^8)-(x^{10})/(10)-1/2 x+C

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integral of (1/(x^9)-x^9-1/2)

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

- \frac{1}{8 x ^{8}} - \frac{x ^{10}}{10} - \frac{1}{2} x + C

Solution steps

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

\int \frac{1}{x ^{9}} d x = - \frac{1}{8 x ^{8}}

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

\int \frac{1}{2} d x = \frac{1}{2} x

= - \frac{1}{8 x ^{8}} - \frac{x ^{10}}{10} - \frac{1}{2} x

Add a constant to the solution

= - \frac{1}{8 x ^{8}} - \frac{x ^{10}}{10} - \frac{1}{2} x + C

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

\int x e^{- \frac{x}{6}} d x

\frac{d}{d x} ((2 x - 5)^{\frac{1}{2}})

\int y^{2} e^{- 2 y} d y

f (x) = x e^{\frac{1}{x}}

What is the integral of (1/(x^9)-x^9-1/2) ?
The integral of (1/(x^9)-x^9-1/2) is -1/(8x^8)-(x^{10})/(10)-1/2 x+C