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

The integral of ((x-2)(x+3))/(x^2) is x+ln|x|+6/x+C

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

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

x + ln ∣ x ∣ + \frac{6}{x} + C

Solution steps

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

Expand

\frac{( x - 2 ) ( x + 3 )}{x ^{2}} : 1 + \frac{1}{x} - \frac{6}{x ^{2}}

= \int 1 + \frac{1}{x} - \frac{6}{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{1}{x} d x - \int \frac{6}{x ^{2}} d x

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x ^{2}} d x = - \frac{6}{x}

= x + ln ∣ x ∣ - (- \frac{6}{x})

Simplify

= x + ln ∣ x ∣ + \frac{6}{x}

Add a constant to the solution

= x + ln ∣ x ∣ + \frac{6}{x} + C

What is the integral of ((x-2)(x+3))/(x^2) ?
The integral of ((x-2)(x+3))/(x^2) is x+ln|x|+6/x+C