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

The integral of x^{-1}(x^{-2}+x^{-1}+2+x^2) is -1/(2x^2)-1/x+2ln|x|+(x^2)/2+C

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

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

- \frac{1}{2 x ^{2}} - \frac{1}{x} + 2 ln ∣ x ∣ + \frac{x ^{2}}{2} + C

Solution steps

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

Simplify

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

= \int \frac{1 + x + 2 x ^{2} + x ^{4}}{x ^{3}} d x

Expand

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

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= - \frac{1}{2 x ^{2}} - \frac{1}{x} + 2 ln ∣ x ∣ + \frac{x ^{2}}{2}

Add a constant to the solution

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

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What is the integral of x^{-1}(x^{-2}+x^{-1}+2+x^2) ?
The integral of x^{-1}(x^{-2}+x^{-1}+2+x^2) is -1/(2x^2)-1/x+2ln|x|+(x^2)/2+C