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

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

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

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

\frac{d}{d x} (tan (11 x))

\int \frac{1}{∣ x ∣ x ^{2} - 1} d x

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

f (y) = 3 y^{2}

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 72 x + 15)

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