What is the integral of x^{-2}(x^6-2x^3+1) ?

The integral of x^{-2}(x^6-2x^3+1) is (x^5)/5-x^2-1/x+C

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integral of x^{-2}(x^6-2x^3+1)

\int x^{- 2} (x^{6} - 2 x^{3} + 1) d x

\frac{x ^{5}}{5} - x^{2} - \frac{1}{x} + C

Solution steps

\int x^{- 2} (x^{6} - 2 x^{3} + 1) d x

Simplify

x^{- 2} (x^{6} - 2 x^{3} + 1) : \frac{x ^{6} - 2 x ^{3} + 1}{x ^{2}}

= \int \frac{x ^{6} - 2 x ^{3} + 1}{x ^{2}} d x

Expand

\frac{x ^{6} - 2 x ^{3} + 1}{x ^{2}} : x^{4} - 2 x + \frac{1}{x ^{2}}

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

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

\int 2 x d x = x^{2}

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

= \frac{x ^{5}}{5} - x^{2} - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{5}}{5} - x^{2} - \frac{1}{x} + C

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

(3 x^{2} y^{2}) d x + (2 x^{3} y + 4 y^{3}) d y = 0

x \to - \frac{π}{3} lim (- 2 cos (x))

\frac{d}{d x} (sin (x^{2}) + cos^{2} (x))

\frac{\partial}{\partial x} (\frac{2}{2 x - y})

What is the integral of x^{-2}(x^6-2x^3+1) ?
The integral of x^{-2}(x^6-2x^3+1) is (x^5)/5-x^2-1/x+C