What is the integral of x^8-4x^{-5}+6 ?

The integral of x^8-4x^{-5}+6 is (x^9)/9+1/(x^4)+6x+C

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integral of x^8-4x^{-5}+6

\int x^{8} - 4 x^{- 5} + 6 d x

\frac{x ^{9}}{9} + \frac{1}{x ^{4}} + 6 x + C

Solution steps

\int x^{8} - 4 x^{- 5} + 6 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^{8} d x - \int 4 x^{- 5} d x + \int 6 d x

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

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

\int 6 d x = 6 x

= \frac{x ^{9}}{9} - (- \frac{1}{x ^{4}}) + 6 x

Simplify

= \frac{x ^{9}}{9} + \frac{1}{x ^{4}} + 6 x

Add a constant to the solution

= \frac{x ^{9}}{9} + \frac{1}{x ^{4}} + 6 x + C

\frac{\partial}{\partial y} (y z^{2})

x \to 0 lim (x^{2} cos (\frac{1}{x}))

f (x) sin (x)

\frac{d y}{d x} = \frac{( 3 x + 4 y )}{( 3 x + 4 y + 4 )}

sin (x) + \frac{7}{3} cot (x)

What is the integral of x^8-4x^{-5}+6 ?
The integral of x^8-4x^{-5}+6 is (x^9)/9+1/(x^4)+6x+C