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

The integral of (x-1/x)^2*x is (x^4)/3-(x^4)/(12)-x^2+ln|x|-1+C

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

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

\frac{x ^{4}}{3} - \frac{x ^{4}}{12} - x^{2} + ln ∣ x ∣ - 1 + C

Solution steps

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

Apply Integration By Parts

: \frac{x ^{4}}{3} - 2 x^{2} - 1 - \int \frac{x ^{3}}{3} - 2 x - \frac{1}{x} d x

= \frac{x ^{4}}{3} - 2 x^{2} - 1 - \int \frac{x ^{3}}{3} - 2 x - \frac{1}{x} d x

\int \frac{x ^{3}}{3} - 2 x - \frac{1}{x} d x = \frac{x ^{4}}{12} - x^{2} - ln ∣ x ∣

= \frac{x ^{4}}{3} - 2 x^{2} - 1 - (\frac{x ^{4}}{12} - x^{2} - ln ∣ x ∣)

Simplify

= \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - x^{2} + ln ∣ x ∣ - 1

Add a constant to the solution

= \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - x^{2} + ln ∣ x ∣ - 1 + C

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What is the integral of (x-1/x)^2*x ?
The integral of (x-1/x)^2*x is (x^4)/3-(x^4)/(12)-x^2+ln|x|-1+C