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RadEn Group TitleBest ResponseYou've already chosen the best response.2
partial fraction 1/(y^21) = A/(y+1) + B/(y1)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
u have to find the values of A and B, first
 one year ago

completeidiot Group TitleBest ResponseYou've already chosen the best response.0
or trig substitution
 one year ago

completeidiot Group TitleBest ResponseYou've already chosen the best response.0
no sorry that would be a terrible idea
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
1/(y^21) = A/(y+1) + B/(y1) 1/(y^21) = ( A(y1) + B(y+1) )/((y+1)(y1)) 1/(y^21) = ((A+B)y + BA ) / (y^21) A+B = 0 BA = 1 by using elimination or subtitution, we get B=1/2 and A = 1/2 see ur integral becomes int (1/2)/(y+1) dy + int(1/2)/(y1) dy
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
use integral by usub
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
for int (1/2)/(y+1) let u = y+1 du = dy it can be : int (1/2)/u du = 1/2 ln(u)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
subtitute back that u = y+1 now, we have int (1/2)/u du = 1/2 ln(u) = 1/2 ln(y+1)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
by same idea to solving int(1/2)/(y1) dy u = y1 du = du so, we have int(1/2)/u = 1/2 ln(u) = 1/2 ln(y1)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
finally, we have the answer : 1/2 ln(y+1) + 1/2 ln(y1) + c
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
does that make sense ?
 one year ago

mash Group TitleBest ResponseYou've already chosen the best response.0
perfect sense ty
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.2
you're welcome
 one year ago
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