What is the solution to the integral (1-y^2/y^2)^2 dy from a calc 2 exam?

[1MileCrash]

Homework Statement

From an exam in calc 2 we are reviewing simple integrals. This one was annoying because it actually contained algebra.. regardless.

[itex]\int(\frac{1 - y^{2}}{y^{2}})^{2} dy[/itex]

Homework Equations

The Attempt at a Solution

First I broke it into two fractions, and turned the second into 1 as it is y squared over y squared.

[itex]\int(\frac{1}{y^{2}} - 1)^{2} dy[/itex]

Then squared the polynomial of sorts.. to get


[itex]\int y^{-4} - y^{-2} - y^{-2} + 1 dy[/itex]

Leading me to a final answer of

[itex]- \frac{y^{-3}}{3} + 2y^{-1} + y + C[/itex]

Look okay? A bit rusty in algebra..

 

[lineintegral1]

Looks fine to me!

 

[Mark44]

Homework Statement

From an exam in calc 2 we are reviewing simple integrals. This one was annoying because it actually contained algebra.. regardless.

[itex]\int(\frac{1 - y^{2}}{y^{2}})^{2} dy[/itex]

Homework Equations

The Attempt at a Solution

First I broke it into two fractions, and turned the second into 1 as it is y squared over y squared.

[itex]\int(\frac{1}{y^{2}} - 1)^{2} dy[/itex]

Then squared the polynomial of sorts.. to get


[itex]\int y^{-4} - y^{-2} - y^{-2} + 1 dy[/itex]

Leading me to a final answer of

[itex]- \frac{y^{-3}}{3} + 2y^{-1} + y + C[/itex]

There's a mistake in your 2nd term. The coefficient of the y^-1 term should be 1, not 2.

Also, a slightly different approach is to square the numerator and denominator of your fraction instead of doing the division first. This leads to the same result, though, so can't really be considered a better approach.

Look okay? A bit rusty in algebra..

 

[1MileCrash]

There's a mistake in your 2nd term. The coefficient of the y^-1 term should be 1, not 2.

How so? After integration of -y^-2 I get +y^-1, and there are two instances of -y^-2. All I did was add them together for 2y^-1.

 

[lineintegral1]

There's a mistake in your 2nd term. The coefficient of the y^-1 term should be 1, not 2.

His integral shows that he is adding y^-2 to y^-2; he just fails to simplify before he integrates. I think that's where his 2 comes from.

 

[Mark44]

Sorry, I totally missed that there was another y^-1 term. My mistake...

 

[1MileCrash]

No problem, thanks guys!