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

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

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

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

\frac{x ^{2}}{2} - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x ^{2} - 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

\frac{d}{d x} (ln (x^{2} + 1) - e^{\frac{x}{2}} cos (π x))

\int \frac{4 x + 5}{x ^{2} + 4 x + 2} d x

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

\frac{d y}{d x} = (\frac{( 2 y + 3 )}{( 4 x + 5 )})^{2}

cos (6 x), 0

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