Find the integral : $\int\frac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$ Find the integral : $\int\dfrac{dx}{x^\frac{1}{2}+x^\frac{1}{3}}$
Please guide which substitution fits in this I am not getting any clue on this .. thanks..
 A: HINT:
As l$cm(2,3)=6$  set  $\displaystyle x^{\dfrac16}=y$
$$\implies x^{\dfrac12}=y^3,x^{\dfrac13}=y^2,x=y^6$$
A: If $x=m^6$, then $dx=6m^5 dm$ and $x^{\frac{1}{2}}=(m^6)^{\frac{1}{2}}=m^3$ and $x^{\frac{1}{3}}=(m^6)^{\frac{1}{3}}=m^2$, then
$$\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\int \frac{6m^5dn}{m^3+m^2}=6\int\frac{m^5dm}{m^2(m+1)}=6\int \frac{m^3dm}{m+1}. $$
Now, $$\frac{m^3}{m+1}=m^2-m+\frac{m}{m+1}, $$ then
$$\int \frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}= 6\int \frac{m^3dm}{m+1}=6\left[\int m^2dm-\int m dm+\int \frac{mdm}{m+1}\right]=$$
$$=6\left[\frac{m^3}{3}-\frac{m^2}{2}+  \int \frac{m+1-1}{m+1}dm \right]= $$
$$=2m^3-3m^2+6\int dm-6\int \frac{dm}{m}=2m^3-3m^2+6m-6\ln |m|+c= $$
$$=2(x^{\frac{1}{6}})^3-3(x^{\frac{1}{6}})^2+6(x^{\frac{1}{6}})-6\ln |x^{\frac{1}{6}}| +c=$$
$$=2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\ln |x^{\frac{1}{6}}| +c. $$
A: After using the substitution posted you'll have to do a bit of work but should get this at your final answer
$$2\sqrt{x}+\ln(\sqrt{x}-1)-\ln(\sqrt{x}+1)-\ln(x-1)-3x^\frac{1}{3}+\ln(x^\frac{2}{3}+x^\frac{1}{3}+1)-2\ln(x^\frac{1}{3}-1)-\ln(x^\frac{1}{3}+x^\frac{1}{6}+1)+2\ln(x^\frac{1}{6}-1)+\ln(x^\frac{1}{3}-x^\frac{1}{6}+1)-2\ln(x^\frac{1}{6}+1)+6x^\frac{1}{6}$$
