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

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

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

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

x - 3 ln ∣ x ∣ - \frac{4}{x} + C

Solution steps

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

Expand

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

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

\int 1 d x = x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

= x - 3 ln ∣ x ∣ - \frac{4}{x}

Add a constant to the solution

= x - 3 ln ∣ x ∣ - \frac{4}{x} + C

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