So the problem is \(\displaystyle \int_0^1\frac{dx}{x^{1/2}+ 8x^{1/3}}\)?
The first thing I would do is see that the least common denominator of those fractional exponents is 6 so write the denominator as \(\displaystyle x^{3/6}+ 8x^{2/6}\) and then decide to let \(\displaystyle u= x^{1/6}\). Then \(\displaystyle x= u^6\) so \(\displaystyle dx= 6u^5du\). When x= 0, u= 0 and when x= 1, u= 1 so the integral becomes \(\displaystyle \int_0^1\frac{6u^5 du}{u^3+ 8u^2}= \int_0^1\frac{6u^3du}{u+ 8}\).
Now, seeing that "u+ 8" in the denominator, I would let v= u+ 8 so that u= v- 8. When u= 0, v= 8 and when u= 1, v= 9 so the integral now is \(\displaystyle \int_8^9 \frac{6(v- 8)^3}{v}dv= \int_8^9 \frac{6v^3- 48v^2+ 192v- 729}{v}dv= \int_8^9 6v^2- 48v+ 192- \frac{729}{v}dv\).
