Worldwide Center of Mathematics - Problem of the Week (u-sub)

\begin{align} u & = \sqrt[6] x \\ u^2 & = \sqrt[3] x \\ u^3 & = \sqrt x \\ u^6 & = x \\ 6u^5\,du & = dx \\[10pt] \int_0^1 \frac 1 {\sqrt{x} + \sqrt[3]{x}} \, dx & = \int_0^1 \frac 1 {u^3 + u^2} \cdot 6u^5 \, du = \int_0^1 \frac 1 {u+1} \cdot 6u^3\,du \\[10pt] & = 6\int_0^1 \left( u^2-u+1 - \frac 1 {u+1} \right) \, du \end{align} The bounds of integration do not change because when $x=0$ then $u=0$ and when $x=1$ then $u=1.$

This is a rationalizing substitution. "Rationalizing" means getting rid of the radicals.

As far as being a limit as $x\downarrow0$ is concerned, that depends to some extent on which of several ways of doing things you're using. Since the function is everywhere nonnegative, if one uses Lebesgue's definition of the integral, then it's not defined differently in such cases, regardless of whether the value of the integral is $+\infty$ or is finite. If you have something like $\displaystyle \int_0^\infty \frac{\sin x} x \, dx,$ where the positive and negative parts are both infinite, then you need to define it as a limit: $\displaystyle \lim_{a\to\infty} \int_0^a \frac{\sin x} x\,dx.$

Yes, the integral is improper. But most of the time it is not a big deal as "evaluating" the antiderivative at $0$ is the same as taking the limit (this requires the functions to be "good enough", but most calculus functions are).

So, taking $x=u^6$, you have $$ dx=6u^5\,du. $$ Then, since the limits don't change (for $x=0,1$ we get $u=0,1$) \begin{align} \displaystyle \int_0^1 \frac 1 {\sqrt{x} + \sqrt[3]{x}} \, dx &=6\int_0^1 \frac {u^5} {u^3 + u^2} \, du =6\int_0^1 \frac {u^3} {u + 1} \, du\\ \ \\ &=6\int_0^1 \frac {u^3+u^2-u^2} {u + 1} \, du\\ \ \\ &=6\int_0^1 \frac {u^3+u^2} {u + 1} \, du-6\int_0^1 \frac {u^2-1+1} {u + 1} \, du\\ &=6\int_0^1 u^2\, du-6\int_0^1(u-1)\,du+6\int_0^1 \frac {1} {u + 1} \, du\\ \ \\ &=2+3-6\log2=5-6\log2 \end{align}