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

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

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

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

\frac{x ^{2}}{2} - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{( x - 1 ) ( x + 1 )}{x} : x - \frac{1}{x}

= \int x - \frac{1}{x} 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 x d x - \int \frac{1}{x} d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - ln ∣ x ∣ + C

\int \frac{1}{a - b x ^{2}} d x

\int 1 + 5 x^{2} d x

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

e^{t}

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