Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com