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Question Number 171171 by Sotoberry last updated on 09/Jun/22

Commented by Rasheed.Sindhi last updated on 09/Jun/22

What to do?

$${What}\:{to}\:{do}? \\ $$

Commented by Sotoberry last updated on 09/Jun/22

integration by partial fraction sir :)

$$\left.{integration}\:{by}\:{partial}\:{fraction}\:{sir}\::\right) \\ $$

Answered by cortano1 last updated on 09/Jun/22

 I=∫ ((x^2 −2x+3)/((x−1)(x^2 −x−1))) dx   = −2ln ∣x−1∣+(3/2)ln ∣x^2 −x−1∣+(1/( (√5)))ln ∣((2x−1−(√5))/(2(√(x^2 −x−1))))∣+c

$$\:{I}=\int\:\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)}\:{dx} \\ $$$$\:=\:−\mathrm{2ln}\:\mid{x}−\mathrm{1}\mid+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} −{x}−\mathrm{1}\mid+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\mathrm{ln}\:\mid\frac{\mathrm{2}{x}−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{1}}}\mid+{c}\: \\ $$

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