What is the integral of ((x^4-1)/x)^2 ?

The integral of ((x^4-1)/x)^2 is (x^7)/7-(2x^3)/3-1/x+C

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integral of ((x^4-1)/x)^2

\int (\frac{x ^{4} - 1}{x})^{2} d x

\frac{x ^{7}}{7} - \frac{2 x ^{3}}{3} - \frac{1}{x} + C

Solution steps

\int (\frac{x ^{4} - 1}{x})^{2} d x

Expand

(\frac{x ^{4} - 1}{x})^{2} : x^{6} - 2 x^{2} + \frac{1}{x ^{2}}

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

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

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

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

= \frac{x ^{7}}{7} - \frac{2 x ^{3}}{3} - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{2 x ^{3}}{3} - \frac{1}{x} + C

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What is the integral of ((x^4-1)/x)^2 ?
The integral of ((x^4-1)/x)^2 is (x^7)/7-(2x^3)/3-1/x+C