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Resolve 2x+3/5(x+2)(2x+1) into partial fractions. |
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Answer» Step-by-step EXPLANATION: Let I= +2x−3 2x 2 +5x−11
= x 2 +2x−3 2(x 2 +2x−3)+x−5
=2+ x 2 +2x−3 x−5
=2+ (x+3)(x−1) x−5
Now (x+3)(x−1) x−5
= (x−1) A
+ (x+3)
⇒x−5=A(x+3)+B(x−1) On COMPARING coefficients we get 1=A+B,−5=3A−B ⇒A=−1,B=2 Therefore I=2− x−1 1
+ x+3 2
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