Question Number 20540 by tammi last updated on 28/Aug/17
$$\int\frac{\sqrt{{x}}}{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx} \\ $$
Answered by $@ty@m last updated on 31/Aug/17
![Let x=t^6 (Remark: LCM of 3 & 2=6) ⇒dx=6t^5 dt ∴∫((√x)/(1+x^(1/3) ))dx=∫((t^3 .6t^5 )/(1+t^2 ))dt =6∫(t^8 /(1+t^2 ))dt =6∫(t^6 −t^4 +t^2 −1+(1/(1+t^2 )))dt (see Note below) =6[(t^7 /7)−(t^5 /5)+(t^3 /3)−t+tan^(−1) t]+C Note: t^6 −t^4 +t^2 −1 −−−−−−−− t^2 +1∣^( ) t^8 t^8 +t^6 −−−− −t^6 −t^6 −t^4 −−−−− t^4 t^4 +t^2 −−−− −t^2 −t^2 −1 −−−− 1](https://www.tinkutara.com/question/Q20681.png)
$${Let}\:{x}={t}^{\mathrm{6}} \:\:\:\:\left({Remark}:\:{LCM}\:{of}\:\mathrm{3}\:\&\:\mathrm{2}=\mathrm{6}\right) \\ $$$$\Rightarrow{dx}=\mathrm{6}{t}^{\mathrm{5}} {dt} \\ $$$$\therefore\int\frac{\sqrt{{x}}}{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx}=\int\frac{{t}^{\mathrm{3}} .\mathrm{6}{t}^{\mathrm{5}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{6}\int\frac{{t}^{\mathrm{8}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{6}\int\left({t}^{\mathrm{6}} −{t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt}\:\left({see}\:{Note}\:{below}\right) \\ $$$$=\mathrm{6}\left[\frac{{t}^{\mathrm{7}} }{\mathrm{7}}−\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}+{tan}^{−\mathrm{1}} {t}\right]+{C}\: \\ $$$${Note}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{6}} −{t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:{t}^{\mathrm{2}} +\mathrm{1}\overset{\:\:} {\mid}{t}^{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{8}} +{t}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{t}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{t}^{\mathrm{6}} −{t}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{4}} +{t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{t}^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$