What is the integral of (x^2+4x^{-2}) ?

The integral of (x^2+4x^{-2}) is (x^3)/3-4/x+C

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integral of (x^2+4x^{-2})

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

\frac{x ^{3}}{3} - \frac{4}{x} + C

Solution steps

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

4 x^{- 2} = \frac{4}{x ^{2}}

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

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

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

= \frac{x ^{3}}{3} - \frac{4}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{4}{x} + C

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What is the integral of (x^2+4x^{-2}) ?
The integral of (x^2+4x^{-2}) is (x^3)/3-4/x+C