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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 $x = 0$?

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    $\begingroup$ $x=u^6$ is the same as $u=x^{1/6}$. $\endgroup$ – Lord Shark the Unknown Dec 18 '17 at 18:18
  • $\begingroup$ this type of substitution is important. It is also used in trigonometric substitution where the new variable is implicitly linked to the old. For example, to solve $\int \sqrt{1-x^2} dx$ one makes a $x = \sin \theta$ substitution. Notice that the substitution is not $\theta = stuff$ so the substition is what I call an "implicit substitution". In contrast, your garden variety $u$-substitutions are what I would term an "explicit substitution". In any event, both answers given are great. $\endgroup$ – James S. Cook Dec 19 '17 at 0:37
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\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: $\displaystyle \lim_{a\to\infty} \int_0^a \frac{\sin x} x\,dx.$

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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}

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  • $\begingroup$ You say it's "improper", but it's not properly improper. I.e. it does not involve both the positive and negative parts begin infinite. $\endgroup$ – Michael Hardy Dec 19 '17 at 6:53
  • $\begingroup$ Yes, I said that because I don't think it is appropriate to talk about Lebesgue integral in a question labeled "calculus" and that talks about "u-sub". $\endgroup$ – Martin Argerami Dec 19 '17 at 11:14
  • $\begingroup$ Is one really talking about the Lebesgue integral, if one says, for example, $$ \int_0^1 \frac{dx}{\sqrt x} = \left[ \vphantom{\frac11} 2\sqrt x \right]_0^1 = 2\sqrt 1 - 2\sqrt 0 = 2 $$ without saying anything about a limit as the lower bound of integration approaches $0\text{ ?} \qquad$ $\endgroup$ – Michael Hardy Dec 19 '17 at 15:35

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