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

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

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

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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Expand

\frac{( x ^{3} - 1 ) ^{2}}{x ^{2}} : x^{4} - 2 x + \frac{1}{x ^{2}}

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

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

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

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

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

Add a constant to the solution

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

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