1.

Resolve `(3x^(2)+7)/(x^(4)-3x^(2)+2)` into partical fractions.A. `(10)/(x^(2)-2)+(5)/(x-1)-(5)/(x+1)`B. `(13)/(x^(2)-2)+(5)/(x+1)-(5)/(x-1)`C. `(5)/(x^(2)-2)+(10)/(x-1)-(10)/(x+1)`D. `(5)/(x-1)-(5)/(x+1)-(13)/(x^(2)-2)`

Answer» Let `x^(2)=p`, then `(3x^(2)+7)/(x^(4)-3x^(2)+2)=(3p+7)/(p^(2)-3p+2)`
Let `(3p+7)/(p^(2)-3p+2)=(A)/(p-1)+(B)/(p-2)`
`implies(3p-7)/((p-1)(p-2))=(A(p-2)+B(p-1))/((p-1)(p-2))`
Consider,
`3p+7=A(p-2)+B(p-1)`
Put `p=1`, `-A=10impliesA=-10`
`p=2`, `B=13`
`(3p+7)/(p^(2)-3p+2)=(13)/(p-2)-(10)/(p-1)`
But `p=x^(2)` ,
`:. (3x^(2)+7)/(x^(4)-3p^(2)+2)=(13)/(x^(2)-2)-(10)/(x^(2)-1)`
`=(13)/(x^(2)-2)-5[(1)/(x-1)-(1)/(x+1)]`
`=(13)/(x^(2)-2)+(5)/(x+1)-(5)/(x-1)`.


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