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mash
 2 years ago
how do you find intergal of 1/(y^2 1)
mash
 2 years ago
how do you find intergal of 1/(y^2 1)

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

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

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

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

RadEn
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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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