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

The integral of ((x+1/x)^2(x-1/x)^2-x+1/x) is (x^5)/5-1/(3x^3)-2x-(x^2)/2+ln|x|+C

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

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

\frac{x ^{5}}{5} - \frac{1}{3 x ^{3}} - 2 x - \frac{x ^{2}}{2} + ln ∣ x ∣ + C

Solution steps

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

\int (x + \frac{1}{x})^{2} (x - \frac{1}{x})^{2} d x = \frac{x ^{5}}{5} - \frac{1}{3 x ^{3}} - 2 x

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{5}}{5} - \frac{1}{3 x ^{3}} - 2 x - \frac{x ^{2}}{2} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{1}{3 x ^{3}} - 2 x - \frac{x ^{2}}{2} + ln ∣ x ∣ + C

5 x^{2} y

x \to 0 lim (\frac{7 x ^{3} - x ^{2}}{x})

(x^{2} + y^{2}) d x - (2 \cdot x \cdot y) d y = 0

x \to 1 - lim (f (x))

\int \frac{x ^{3} + 5}{x ^{2} + 3 x + 2} d x

What is the integral of ((x+1/x)^2(x-1/x)^2-x+1/x) ?
The integral of ((x+1/x)^2(x-1/x)^2-x+1/x) is (x^5)/5-1/(3x^3)-2x-(x^2)/2+ln|x|+C