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

\frac{\partial}{\partial y} (arctan (x t))

\frac{\partial}{\partial x} (\frac{sin ( x )}{sin ( y )})

\int_{0}^{4} (4 - x) x d x

\int_{0}^{4} π (\frac{1}{x + 3})^{2} d x

\int_{0}^{x} t^{2} cos (t) d t

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