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Question Number 158176 by Odhiambojr last updated on 31/Oct/21

∫{((x^2 −x−21)/(2x^3 −x^2 +8x−4))}dx

$$\int\left\{\frac{{x}^{\mathrm{2}} −{x}−\mathrm{21}}{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{4}}\right\}{dx}\: \\ $$

Answered by MJS_new last updated on 31/Oct/21

=∫((x^2 −x−21)/((2x−1)(x^2 +4)))dx=  =∫(((3x)/(x^2 +4))−(5/(2x−1))+(1/(x^2 +4)))dx=  =(3/2)ln (x^2 +4) −(5/2)ln ∣2x−1∣ +(1/2)arctan (x/2) +C

$$=\int\frac{{x}^{\mathrm{2}} −{x}−\mathrm{21}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}= \\ $$$$=\int\left(\frac{\mathrm{3}{x}}{{x}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{5}}{\mathrm{2}{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}\right){dx}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}\right)\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{2}{x}−\mathrm{1}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\frac{{x}}{\mathrm{2}}\:+{C} \\ $$

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