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

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

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integral of (4x^{-2}-3)

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

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

Solution steps

\int 4 x^{- 2} - 3 d x

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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