Question Number 116564 by bemath last updated on 05/Oct/20
$$\int\:\frac{\mathrm{dx}}{\mathrm{x}+\sqrt[{\mathrm{3}\:}]{\mathrm{x}}}\:? \\ $$
Answered by MJS_new last updated on 05/Oct/20
![∫(dx/(x^(1/3) (x^(2/3) +1)))= [t=x^(2/3) +1 → dx=(3/2)x^(1/3) dt] =(3/2)∫(dt/t)=(3/2)ln t =(3/2)ln (x^(2/3) +1) +C](https://www.tinkutara.com/question/Q116567.png)
$$\int\frac{{dx}}{{x}^{\mathrm{1}/\mathrm{3}} \left({x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}\right)}= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}\:\rightarrow\:{dx}=\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{1}/\mathrm{3}} {dt}\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dt}}{{t}}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:{t}\:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}\right)\:+{C} \\ $$
Commented by peter frank last updated on 05/Oct/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by bobhans last updated on 05/Oct/20
$$\mathrm{another}\:\mathrm{way} \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\lambda^{\mathrm{3}} \:\Rightarrow\mathrm{dx}\:=\:\mathrm{3}\lambda^{\mathrm{2}} \:\mathrm{d}\lambda \\ $$$$\int\:\frac{\mathrm{3}\lambda^{\mathrm{2}} }{\lambda^{\mathrm{3}} +\lambda}\:\mathrm{d}\lambda\:=\:\int\:\frac{\mathrm{3}\lambda}{\lambda^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}\lambda\:=\:\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{\mathrm{d}\left(\lambda^{\mathrm{2}} +\mathrm{1}\right)}{\lambda^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\left(\lambda^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{C}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} \:}\:+\mathrm{1}\right)\:+\:\mathrm{C} \\ $$
Commented by peter frank last updated on 05/Oct/20
$$\mathrm{thank}\:\mathrm{you} \\ $$