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Indefinite Integration

Section: Mathematics

Question : 2 of 89

Marks: +1, -0

\;\frac{3x^3-7x+1}{(x-2)^5} = \;\frac{A}{x-2} + \;\frac{B}{(x-2)^2} + \;\frac{C}{(x-2)^3} + \;\frac{D}{(x-2)^4} + \;\frac{E}{(x-2)^5}

A(B+C+D+E)=

Solution:

Given,

\;\frac{3x^3-7x+1}{(x-2)^5}

\;=\;\frac{A}{x-2}+\;\frac{B}{(x-2)^2}+\;\frac{C}{(x-2)^3}+\;\frac{D}{(x-2)^4}+\;\frac{E}{(x-2)^5}

\;=\;\frac{A(x-2)^4+B(x-2)^3+C(x-2)^2+D(x-2)+E}{(x-2)^5}

\;=A x^4+(-8 A+B) x^3+(24 A-6 B+C) x^2+(-32 A+12 B-4 C+D) x+16 A-8 B+4 C-2 D+E

Comparing the coefficient, we get

\;A=0

\;-8 A+B=3

\; \Rightarrow \; \; B=3+8 A=3

\;24 A-6 B+C=0

\; \Rightarrow \; \; C=6 B-24 A=18-0=18

\;-32 A+12 B-4 C+D=-7

\; \Rightarrow \; \; D=-7+32 A-12 B+4 C=-7+0-36+72=29

\;16 A-8 B+4 C-2 D+E=1

\; \Rightarrow \; \; E=1-16 A+8 B-4 C+2 D

\;=1-0+24-72+58=11

\; \therefore \; \; A(B+C+D+E)=0(3+18+29+11)=0

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