What is the integral of (x-1/x)^3 ?

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

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integral of (x-1/x)^3

\int (x - \frac{1}{x})^{3} d x

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

Solution steps

\int (x - \frac{1}{x})^{3} d x

Expand

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

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

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

\int 3 x d x = \frac{3 x ^{2}}{2}

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

Simplify

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

Add a constant to the solution

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

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

x \to - 10 lim (7)

\int (y cos (x y)) d x

x \to - 7 + lim (\frac{x + 8}{x + 7})

\frac{d}{d x} (5 ln (4 x^{2}))

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