What is the integral of ((x+1)(x-2))/(x^2) ?

The integral of ((x+1)(x-2))/(x^2) is ((x+1)(x-2))/x-ln|x|+4/x+C

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integral of ((x+1)(x-2))/(x^2)

\int \frac{( x + 1 ) ( x - 2 )}{x ^{2}} d x

\frac{( x + 1 ) ( x - 2 )}{x} - ln ∣ x ∣ + \frac{4}{x} + C

Solution steps

\int \frac{( x + 1 ) ( x - 2 )}{x ^{2}} d x

Apply Integration By Parts

: \frac{( x + 1 ) ( x - 2 )}{x} - \int \frac{x + 4}{x ^{2}} d x

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

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

= \frac{( x + 1 ) ( x - 2 )}{x} - (ln ∣ x ∣ - \frac{4}{x})

Simplify

= \frac{( x + 1 ) ( x - 2 )}{x} - ln ∣ x ∣ + \frac{4}{x}

Add a constant to the solution

= \frac{( x + 1 ) ( x - 2 )}{x} - ln ∣ x ∣ + \frac{4}{x} + C

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What is the integral of ((x+1)(x-2))/(x^2) ?
The integral of ((x+1)(x-2))/(x^2) is ((x+1)(x-2))/x-ln|x|+4/x+C