What is the integral of (x^2-1/(x^2))^3 ?

The integral of (x^2-1/(x^2))^3 is (x^7)/7-x^3-3/x+1/(5x^5)+C

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integral of (x^2-1/(x^2))^3

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

\frac{x ^{7}}{7} - x^{3} - \frac{3}{x} + \frac{1}{5 x ^{5}} + C

Solution steps

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

Expand

(x^{2} - \frac{1}{x ^{2}})^{3} : x^{6} - 3 x^{2} + \frac{3}{x ^{2}} - \frac{1}{x ^{6}}

= \int x^{6} - 3 x^{2} + \frac{3}{x ^{2}} - \frac{1}{x ^{6}} 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^{6} d x - \int 3 x^{2} d x + \int \frac{3}{x ^{2}} d x - \int \frac{1}{x ^{6}} d x

\int x^{6} d x = \frac{x ^{7}}{7}

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

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

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

= \frac{x ^{7}}{7} - x^{3} - \frac{3}{x} - (- \frac{1}{5 x ^{5}})

Simplify

= \frac{x ^{7}}{7} - x^{3} - \frac{3}{x} + \frac{1}{5 x ^{5}}

Add a constant to the solution

= \frac{x ^{7}}{7} - x^{3} - \frac{3}{x} + \frac{1}{5 x ^{5}} + C

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What is the integral of (x^2-1/(x^2))^3 ?
The integral of (x^2-1/(x^2))^3 is (x^7)/7-x^3-3/x+1/(5x^5)+C