Question Number 211152 by Spillover last updated on 29/Aug/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\sqrt{{x}}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{dx} \\ $$$$ \\ $$
Answered by Ghisom last updated on 30/Aug/24
![∫((√x)/(x^3 +x^2 +x+1))dx=∫((√x)/((x+1)(x^2 +1)))dx= [t=(√x) → dx=2(√x)dt] =2∫(t^2 /((t^2 +1)(t^4 +1)))dt= =−∫(dt/(t^2 +1))+(1/2)(∫(dt/(t^2 −(√2)t+1))+∫(dt/(t^2 +(√2)t+1)))= =−arctan t +((√2)/2)(arctan ((√2)t−1) +arctan ((√2)t+1))= =−arctan (√x) +((√2)/2)(arctan ((√(2x))−1) +arctan ((√(2x))+1) +C](https://www.tinkutara.com/question/Q211154.png)
$$\int\frac{\sqrt{{x}}}{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\int\frac{\sqrt{{x}}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}{dt}= \\ $$$$=−\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\int\frac{{dt}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\int\frac{{dt}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}\right)= \\ $$$$=−\mathrm{arctan}\:{t}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\right)= \\ $$$$=−\mathrm{arctan}\:\sqrt{{x}}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}{x}}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}{x}}+\mathrm{1}\right)\:+{C}\right. \\ $$