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

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

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

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

RadEn
 one year ago
Best ResponseYou've already chosen the best response.21/(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

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

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

RadEn
 one year ago
Best ResponseYou've already chosen the best response.2by 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)

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