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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 ( y ^{2} - 1 ) ( y ^{2} + 1 )}{y ^{2}} d y = 2 (- \frac{( y ^{2} - 1 ) ( y ^{2} + 1 )}{y} + \frac{4 y ^{3}}{3})

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

Add a constant to the solution

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

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