x⁴ - x²+1/ x²(x²-1)² resolve into partial fractions
Answers
Step-by-step explanation:
We have
x
4
+x
2
+1
x
=
(x
2
+x+1)(x
2
−x+1)
x
(Remember)
Let
(x
2
+x+1)(x
2
−x+1)
x
=
x
2
+x+1
Ax+B
+
x
2
−x+1
Cx+D
...(i)
⟹x=(Ax+B)(x
2
−x+1)+(Cx+D)(x
2
+x+1) ...(ii)
Putting x
2
−x+1=0
or x
2
=x−1 in (ii), we obtain
x=0+(Cx+D)(x−1+x+1)
⟹x=(Cx+D)2x
⟹
2
1
=Cx+D
On comparing, C=0 and D=
2
1
Again putting x
2
+x+1=0
or x
2
=−x−1 in (ii), we obtain
x=(Ax+B)(−x−1−x+1)+0
⟹x=(Ax+B)(−2x)
⟹−
2
1
=Ax+B
On Comparing, A=0,B=−
2
1
Substituting the value of A, B, C, D in (i), we have
(x
2
+x+1)(x
2
−x+1)
x
=
x
2
+x+1
0−
2
1
+
x
2
−x+1
0+
2
1
or
x
4
+x
2
+1
x
=−
2(x
2
+x+1)
1
+
2(x
2
−x+1)
1
which are the required partial fractions.