Question Number 33690 by mondodotto@gmail.com last updated on 22/Apr/18
$$\int\frac{\boldsymbol{\mathrm{dx}}}{\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}}} \\ $$
Answered by math1967 last updated on 22/Apr/18
![let x=z^6 ∴dx=6z^5 dz 6∫((z^5 dz)/(z^2 +z^3 ))=6∫((z^5 dz)/(z^2 (1+z)))=6∫((z^3 dz)/(1+z)) =6∫(((z^3 +1)dz)/(z+1)) −6∫(dz/(z+1)) =6∫(z^2 −z+1)dz −6ln[z+1] =6(z^3 /3) −((6z^2 )/2) +6z −6ln[z+1] +c =2(√(x )) −3x^(1/3) +6x^(1/6) −6ln[x^(1/6) +1] +c](Q33696.png)
$${let}\:{x}={z}^{\mathrm{6}} \:\:\therefore{dx}=\mathrm{6}{z}^{\mathrm{5}} {dz} \\ $$$$\mathrm{6}\int\frac{{z}^{\mathrm{5}} {dz}}{{z}^{\mathrm{2}} +{z}^{\mathrm{3}} }=\mathrm{6}\int\frac{{z}^{\mathrm{5}} {dz}}{{z}^{\mathrm{2}} \left(\mathrm{1}+{z}\right)}=\mathrm{6}\int\frac{{z}^{\mathrm{3}} {dz}}{\mathrm{1}+{z}} \\ $$$$=\mathrm{6}\int\frac{\left({z}^{\mathrm{3}} +\mathrm{1}\right){dz}}{{z}+\mathrm{1}}\:−\mathrm{6}\int\frac{{dz}}{{z}+\mathrm{1}} \\ $$$$=\mathrm{6}\int\left({z}^{\mathrm{2}} −{z}+\mathrm{1}\right){dz}\:−\mathrm{6}{ln}\left[{z}+\mathrm{1}\right] \\ $$$$=\mathrm{6}\frac{{z}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{\mathrm{6}{z}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{6}{z}\:−\mathrm{6}{ln}\left[{z}+\mathrm{1}\right]\:+{c} \\ $$$$=\mathrm{2}\sqrt{{x}\:\:}\:\:−\mathrm{3}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\mathrm{6}{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \:−\mathrm{6}{ln}\left[{x}^{\frac{\mathrm{1}}{\mathrm{6}}} +\mathrm{1}\right]\:+{c} \\ $$