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15. Integration Using Partial Fractions
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Q28 of 168 Page 762

Evaluate:



Let

Now putting,


A(x+1)2+B(x+2)(x+1)+C(x+2)=x2+x+1


Now put x+1=0


Therefore, x=-1


A(0)+B(0)+C(-1+2) =1-1+1=1


C=1


Now put x+2=0


Therefore, x=-2


A(-2+1)2+B(0)+C(0) =4-2+1=3


A=3


Equating the coefficient of x2,A+B=1


3+B=1


B=-2


Form equation (1),we get,



So,




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Questions · 168
15. Integration Using Partial Fractions
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