[m]\sqrt[6]{x}=t[/m] ⇒ [m]\sqrt[3]{x}=t^2[/m] и [m]\sqrt[2]{x}=t^3[/m]
[m]x=t^6[/m]
[m]dx=6t^5dt[/m]
получаем интеграл:
[m] ∫= \frac{6t^5}{t^3(1+t^2)}dt= 6∫ \frac{t^2}{1+t^2}dt=[/m]
Выделяем целую часть
[m]= 6∫ \frac{(t^2+1-1)}{1+t^2}dt=6∫ \frac{(t^2+1)dt}{1+t^2}dt-6∫ \frac{1}{1+t^2}dt=[/m]
[m]=6 ∫dt-6 ∫ \frac{1}{1+t^2}dt=6t-6arctgt+C=6\sqrt[6]{x}-6arctg\sqrt[6]{x}+C[/m]