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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

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