Question-117143

Question Number 117143 by aurpeyz last updated on 09/Oct/20

Answered by prakash jain last updated on 13/Oct/20

((2x^4 −2x^3 +4x^2 −2x)/((x−1)(x^2 +1))) =Ax+B+(C/(x−1))+((Dx+E)/(x^2 +1)) 2x^4 −2x^3 +4x^2 −2x= (Ax+B)(x−1)(x^2 +1)+C(x^2 +1)+(Dx+E)(x−1) coefficient of x^4 ⇒A=2 put x=0 ⇒−B+C−E=0 put x=1⇒2C=2⇒C=1 put x=i⇒2+2i−4−2i=(Di+E)(i−1) −2=−(E+D)+i(E−D) ⇒E=D=1 −B+C−E=0⇒B=0 A=2,B=0,C=1,D=1,E=1 2x+(2/(x−1))+((x+1)/(x^2 +1))

$$\frac{\mathrm{2}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$={Ax}+{B}+\frac{{C}}{{x}−\mathrm{1}}+\frac{{Dx}+{E}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{2}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}= \\ $$$$\:\:\left({Ax}+{B}\right)\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{C}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({Dx}+{E}\right)\left({x}−\mathrm{1}\right) \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{4}} \:\:\Rightarrow{A}=\mathrm{2} \\ $$$${put}\:{x}=\mathrm{0}\:\Rightarrow−{B}+{C}−{E}=\mathrm{0}\:\:\: \\ $$$${put}\:{x}=\mathrm{1}\Rightarrow\mathrm{2}{C}=\mathrm{2}\Rightarrow{C}=\mathrm{1} \\ $$$${put}\:{x}={i}\Rightarrow\mathrm{2}+\mathrm{2}{i}−\mathrm{4}−\mathrm{2}{i}=\left({Di}+{E}\right)\left({i}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:−\mathrm{2}=−\left({E}+{D}\right)+{i}\left({E}−{D}\right) \\ $$$$\:\:\:\:\:\:\Rightarrow{E}={D}=\mathrm{1} \\ $$$$−{B}+{C}−{E}=\mathrm{0}\Rightarrow{B}=\mathrm{0} \\ $$$${A}=\mathrm{2},{B}=\mathrm{0},{C}=\mathrm{1},{D}=\mathrm{1},{E}=\mathrm{1} \\ $$$$\mathrm{2}{x}+\frac{\mathrm{2}}{{x}−\mathrm{1}}+\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$

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