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Question

Resolve x4x2+1x2(x2+1)2 into partial fractions.

A
2x2+1(x2+1)2
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B
1x2+3(x2+1)2
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C
1x25(x2+1)2
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D
1x23(x2+1)2
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Solution

The correct option is D 1x23(x2+1)2
The given fraction has x2 everywhere then put x2=t for the sake of partial fractions.
x4x2+1x2(x2+1)2=t2t+1t(t+1)2
Now let t2t+1t(t+1)2=At+Bt+1=C(t+1)2 ...(i)
t2t+1=A(t+1)2+Bt(t+1)+Ct ...(ii)
Substituting t+1=0 or t=1 in (ii), we ontain
(1)2(1)+1=0+0+C(1)
C=3
Substituting t=0 in (ii) we obtain
00+1=A(0+1)2+0+0
A=1
Now equating the coefficient of t2 on both sides of (ii) then
1=A+B
1=1+B
B=0
Substituting the values of A, B and C in (i) we get
t2t+1t(t+1)2=1t3(t+1)2
In last substitution t=x2
then x4x2+1x2(x2+1)2=1x23(x2+1)2
which are the required partial fractions.
Hence, option 'D' is correct.

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