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Question

can someone help me with integrating (y/(3-y))^2? I have no idea where to start and the answer key is confusing the heck out of me.

Please help!

Renie · Accepted Answer

Y^2/(y^2-6y +9)  is an improper fraction for integration: the numerator should be lower degree than the denominator: so rewrite using long division or regrouping:

1+  (6y -9)/(y^2-6y+9)... Notice the derivative of the denominator is 2y -6, so rewrite 6y-9= 3(2y-6)+ 9

INT[  1 +(3) (2y-6)/(y^2-6y + 9) + (9)/[(y-3)^2] dy

= y + 3* ln(y-3)^2 -  9(y-3)^-1+ C

I hope this helps!

Anónimo · Answer

y^2/(3-y)^2 = y^2 / (y^2 - 6y + 9) = 1 + (6y - 9) / (y^2 - 6y + 9)
= 1 + 3 * (2y - 6) / (y^2 - 6y + 9) + 9 / (y^2 - 6y + 9)
= 1 + 3* (2y - 6) / (y^2 - 6y + 9) + 9 / (y - 3)^2

So, now integrate:

y + 3*log(|y-3|^2) - 9 / (y-3) + c
= y + 6*log(|y-3|) - 9 / (y-3) + c

Hope this helps.

JoAn · Answer

Evaluate: â«(y/(3-y))Â²dy

let

z = 3 - y
y = 3 - z

dz = -dy
dy = -dz

thus

â«(y/(3-y))Â²dx = â«[((3-z)/z)Â²](-dz)

â«(y/(3-y))Â²dx = -â«[((3-z)Â²/zÂ²)]dz

â«(y/(3-y))Â²dx = -â«[(9 - 6z + zÂ²)/zÂ²]dz

â«(y/(3-y))Â²dx = -â«[9/zÂ² - 6z/zÂ² + zÂ²/zÂ²]dz

â«(y/(3-y))Â²dx = -â«[9/zÂ² - 6/z + 1]dz

â«(y/(3-y))Â²dx = -â��[9/zÂ²]dz - â«[-6/z]dz - â«[1]dz

â«(y/(3-y))Â²dx = -9â«[1/zÂ²]dz + 6â«[1/z]dz - â«dz

â«(y/(3-y))Â²dx = -9(-1)(1/z) + 6(ln z) - z

â«(y/(3-y))Â²dx = 9/z + 6(ln z) - z

â«(y/(3-y))Â²dx = 9/z - z + ln z⁶

â«(y/(3-y))Â²dx = (9 - z)Â²/z + ln z⁶

recall z = 3 - y

â«(y/(3-y))Â²dx = (9 - (3 - y))Â²/(3 - y) + ln (3 - y)⁶

â«(y/(3-y))Â²dx = (6 + y)Â²/(3 - y) + ln (3 - y)⁶  <---------- answer

â«(y/(3-y))Â²dx = (6 + y)Â²/(3 - y) + ln (3 - y)⁶

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