TITLE: Worldwide Center of Mathematics - Problem of the Week (u-sub)
QUESTION [6 upvotes]: The problem is to integrate $\displaystyle \int_0^1 \frac 1 {\sqrt{x} + \sqrt[3]{x}} \, dx$ and is solved by using the substitution $x = u^6$. The way I learned u-sub one has to find an expression for $u$ (i.e. $u =$ [expression in terms of $x$]) and not for $x$. The explanation in the solution is short and I do not quite understand this type of u-sub.
Also, should this integral not be a limit, since the original function is undefined for $x = 0$?

REPLY [6 votes]: 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}