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

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

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

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

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

Solution steps

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

Expand

(x^{2} - \frac{1}{x ^{2}})^{2} : x^{4} - 2 + \frac{1}{x ^{4}}

= \int x^{4} - 2 + \frac{1}{x ^{4}} 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^{4} d x - \int 2 d x + \int \frac{1}{x ^{4}} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 2 d x = 2 x

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

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

Add a constant to the solution

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

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