Evaluate:
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Let ![]()
Which implies ![]()
Therefore, I=x+I1
Where, I1=![]()
Putting ![]()
Which implies,
A(x+2)+B(x+1)=2x+1
Now put x+2=0
Therefore, x=-2
A(0)+B(-1)=-4+1
B=3
Now put x+1=0
Therefore, x=-1
A(-1+2)+B(0)=-2+1
A=-1
Now From equation (1) we get,
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