Question-137461

Question Number 137461 by mathlove last updated on 03/Apr/21

Answered by Ñï= last updated on 03/Apr/21

∫((1+x^2 )/(2x+x^3 ))dx=(1/3)∫((2+3x^2 +1)/(2x+x^3 ))dx=(1/3)ln∣2x+x^3 ∣+(1/3)∫(dx/(x(2+x^2 ))) =(1/3)ln∣2x+x^3 ∣+(1/6)∫((1/x)−(x/(2+x^2 )))=(1/3)ln∣2x+x^3 ∣+(1/6)ln∣x∣−(1/(12))ln∣2+x^2 ∣+C

$$\int\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}+{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{2}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}+{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{2}{x}+{x}^{\mathrm{3}} \mid+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{2}{x}+{x}^{\mathrm{3}} \mid+\frac{\mathrm{1}}{\mathrm{6}}\int\left(\frac{\mathrm{1}}{{x}}−\frac{{x}}{\mathrm{2}+{x}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{2}{x}+{x}^{\mathrm{3}} \mid+\frac{\mathrm{1}}{\mathrm{6}}{ln}\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{12}}{ln}\mid\mathrm{2}+{x}^{\mathrm{2}} \mid+{C} \\ $$

Answered by MJS_new last updated on 03/Apr/21

∫((x^2 +1)/(x(x^2 +2)))dx=(1/2)∫(dx/x)+(1/2)∫(x/(x^2 +2))= =(1/2)ln ∣x∣ +(1/4)ln (x^2 +2) +C

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

Answered by liberty last updated on 04/Apr/21

∫ ((x^2 +1)/(x(2+x^2 ))) dx = ∫ (x/(2+x^2 ))dx+∫(1/(x(2+x^2 )))dx =(1/2)∫ ((d(2+x^2 ))/(2+x^2 ))+∫(((√2) sec^2 t)/( (√2) tan t(2sec^2 t)))dt [ x=(√2) tan t ] I=(1/2)ln (2+x^2 )+(1/2)∫((cos t)/(sin t)) dt I=(1/2)ln (2+x^2 )+(1/2)ln (sin t)+c I=(1/2)ln (2+x^2 )+(1/2)ln (((∣x∣)/( (√(2+x^2 )))))+c

$$\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}\:{dx}\:=\:\int\:\frac{{x}}{\mathrm{2}+{x}^{\mathrm{2}} }{dx}+\int\frac{\mathrm{1}}{{x}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\mathrm{2}+{x}^{\mathrm{2}} }+\int\frac{\sqrt{\mathrm{2}}\:\mathrm{sec}\:^{\mathrm{2}} {t}}{\:\sqrt{\mathrm{2}}\:\mathrm{tan}\:{t}\left(\mathrm{2sec}\:^{\mathrm{2}} {t}\right)}{dt} \\ $$$$\:\left[\:{x}=\sqrt{\mathrm{2}}\:\mathrm{tan}\:{t}\:\right] \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}+{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cos}\:{t}}{\mathrm{sin}\:{t}}\:{dt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}+{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{sin}\:{t}\right)+{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}+{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mid{x}\mid}{\:\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}\right)+{c} \\ $$

Commented by MJS_new last updated on 04/Apr/21

(1/2)ln a +(1/2)ln (b/( (√a))) =(1/2)ln a +(1/2)ln b −(1/2)ln (√a) = =(1/2)ln a +(1/2)ln b −(1/4)ln a =(1/4)ln a +(1/2)ln b

$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{a}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{b}}{\:\sqrt{{a}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{a}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{b}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\sqrt{{a}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{a}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{b}\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{a}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{a}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{b} \\ $$

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