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

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

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

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

\frac{x ^{4}}{4} - x^{2} + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

\int x^{3} d x = \frac{x ^{4}}{4}

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{4}}{4} - x^{2} + ln ∣ x ∣

Add a constant to the solution

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

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