[tex]\displaystyle \int \dfrac{dx}{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}} ;\\\text{Mai intai dam factor fortat un }\sqrt[4]{x} \text{ de la numitor(o sa vezi de ce)}.\\= \int \dfrac{dx}{\sqrt[4]{x}(\sqrt[4]{x}+\sqrt[12]{x}+2)}\\\text{Incercam sa descompunem paranteza de la numitor:}\\\sqrt[12]{x}+\sqrt[4]{x}+2 \stackrel{\sqrt[12]{x}=t}{~~=} t^3+t+2=t^3+t^2-t^2-t+2t+2=\\=t^2(t+1)-t(t+1)+2(t+1)=(t+1)(t^2-t+2)=\\(\sqrt[12]{x}+1)(\sqrt[6]{x}-\sqrt[12]{x}+2)[/tex]
[tex]\text{Mai putem face inca o jmecherie.Profitam de faptul ca }\\\sqrt[4]{x}=x^{\frac{1}{4}}= x^ {\frac{11}{12}-\frac{2}{3}}=\dfrac{x^{\frac{11}{12}}}{x^{\frac{2}{3}}},\text{ deci } \dfrac{1}{\sqrt[4]{x}}=\dfrac{x^{\frac{2}{3}}}{x^{\frac{11}{12}}}\\ \text{Cu ce ne-a ajutat asta ?In primul rand }x^{\frac{2}{3}}=(\sqrt[12]{x})^8,\text{iar }x^{\frac{11}{12}} \text{ ne va}\\\text{ajuta la schimbarea de variabila.}\\\sqrt[12]{x}=t\Rightarrow dx=12x^{\frac{11}{12}}dt\\\text{Revenind,integrala devine :}[/tex]
[tex]=\displaystyle\int \dfrac{u^8}{(u+1)(u^2-u+2)}du\\\text{De aici trebuie neaparat sa impartim polinoamele. }\\\text{Voi sari peste partea aia.}\\=\int \left(u^5-u^3-2u^2+u+4+\dfrac{3u^2-6u-8}{(u+1)(u^2-u+2)}\right)du\\\text{De aici cred ca te descurci.}[/tex]
