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Question Number 4502 by #na last updated on 02/Feb/16

(4/(x−1))−((2x)/(x^2 +1))−(2/(x+1))+(6/(1−x^2 ))

$$\frac{\mathrm{4}}{{x}−\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}+\frac{\mathrm{6}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$

Answered by Rasheed Soomro last updated on 02/Feb/16

(4/(x−1))−((2x)/(x^2 +1))−(2/(x+1))−(6/((x−1)(x+1))) ((4(x+1)(x^2 +1)−2x(x−1)(x+1)−2(x−1)(x^2 +1)−6(x^2 +1))/((x−1)(x+1)(x^2 +1))) ((4(x^3 +x^2 +x+1)−2x(x^2 −1)−2(x^3 −x^2 +x−1)−6(x^2 +1))/((x−1)(x+1)(x^2 +1))) ((4x^3 +4x^2 +4x+4−2x^3 +2x−2x^3 +2x^2 −2x+2−6x^2 −6)/((x−1)(x+1)(x^2 +1))) =((4x)/((x−1)(x+1)(x^2 +1)))

$$\frac{\mathrm{4}}{{x}−\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}−\frac{\mathrm{6}}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{4}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}{x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)−\mathrm{2}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{6}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{\mathrm{4}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}−\mathrm{1}\right)−\mathrm{6}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}{x}}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$ \\ $$

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