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Mathematics 17 Online
OpenStudy (anonymous):

Integration.

rvc (rvc):

subsitution method

hartnn (hartnn):

write \(z^4 = (z^4 -1)+1\)

hartnn (hartnn):

then factor z^4 -1

rvc (rvc):

y like that @hartnn

hartnn (hartnn):

because denominator z^2 -1 is one of the factors of z^4-1

OpenStudy (anonymous):

yes @hartnn

hartnn (hartnn):

so you will simply get (z^2-1)(z^2+1) +1 in the numerator, right ?? now just separate out the numerators :)

OpenStudy (anonymous):

what do you mean by separate out..

hartnn (hartnn):

like (a+b)/c = a/c +b/c

OpenStudy (anonymous):

im not sure.. but. (z^2 -1 + 1) (z^2 + 1 +1 )

hartnn (hartnn):

|dw:1404542108621:dw| got that ^^ ?

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