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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

Add a constant to the solution

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

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