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

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

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integral of (3x^2-4)/(x^2)

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

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

Solution steps

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

Expand

\frac{3 x ^{2} - 4}{x ^{2}} : 3 - \frac{4}{x ^{2}}

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

\int 3 d x = 3 x

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

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

Simplify

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

Add a constant to the solution

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

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