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Question Number 208316 by Tawa11 last updated on 11/Jun/24
∫ ((x^2 + 3)/(x^2 (x + 1)(x^2 + 1)^2 )) dx
$$\int\:\frac{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{3}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}\:\:+\:\:\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Answered by Frix last updated on 11/Jun/24
∫((x^2 +3)/(x^2 (x+1)(x^2 +1)^2 ))dx= =−3∫(dx/x)+ =−3ln ∣x∣ − +3∫(dx/x^2 )+ −(3/x)+ +∫(dx/(x+1))+ +ln ∣x+1∣ + +∫((2x)/(x^2 +1))dx− +ln (x^2 +1) − −(5/2)∫(dx/(x^2 +1))+ −(5/2)tan^(−1) x − +(1/2)∫((x^2 +2x−1)/((x^2 +1)^2 ))dx −((x+1)/(2(x^2 +1)))= =−(((7x^2 +x+6))/(2x(x^2 +1)))+ln ∣(((x+1)(x^2 +1))/x^3 )∣ −(5/2)tan^(−1) x +C
$$\int\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}= \\ $$$$=−\mathrm{3}\int\frac{{dx}}{{x}}+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{3ln}\:\mid{x}\mid\:− \\ $$$$+\mathrm{3}\int\frac{{dx}}{{x}^{\mathrm{2}} }+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{3}}{{x}}+ \\ $$$$+\int\frac{{dx}}{{x}+\mathrm{1}}+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:+ \\ $$$$+\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:− \\ $$$$−\frac{\mathrm{5}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:{x}\:− \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:\:\:\:−\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}= \\ $$$$=−\frac{\left(\mathrm{7}{x}^{\mathrm{2}} +{x}+\mathrm{6}\right)}{\mathrm{2}{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\mathrm{ln}\:\mid\frac{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{3}} }\mid\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:{x}\:+{C} \\ $$
Commented by Frix last updated on 11/Jun/24
(1/2)∫((x^2 +2x−1)/((x^2 +1)^2 ))dx =^(t=tan^(−1) x) =∫((1/2)+cos t sin t −cos^2 t)dt= =−((√2)/2)∫cos (2t+(π/4)) dt=−((√2)/4)sin (2t+(π/4)) = =((x^2 −2x−1)/(4(x^2 +1)))=(1/4)−((x+1)/(2(x^2 +1)))=−((x+1)/(2(x^2 +1)))+C
$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\overset{{t}=\mathrm{tan}^{−\mathrm{1}} \:{x}} {=} \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}\:−\mathrm{cos}^{\mathrm{2}} \:{t}\right){dt}= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\mathrm{cos}\:\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{4}}\right)\:{dt}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{sin}\:\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{4}}\right)\:= \\ $$$$=\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}}{\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=−\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+{C} \\ $$
Commented by Tawa11 last updated on 11/Jun/24
Thanks sir. I really appreciate sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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