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I'm solving integrals and the integral is a rational function with these two radicals at the bottom i.e. $\sqrt{x}$ + $\sqrt[5]{x^2}$.

The integral is $\int\frac{dx}{x(\sqrt{x} + \sqrt[5]{x^2})} $

The way this integral is solved, it says that I first must represent these both radicals as $t$. How to get a $t$ from both of them?

In the answers it says that $x=t^{10}$ i.e. $t=\sqrt[10]{x}$

Here's an image of the integral and the answer.

BoostedAnimal
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    This question makes no sense. Please rephrase. – K.defaoite Sep 05 '20 at 18:39
  • You have two roots ... roots of what equation? For what purpose do you want to express $x$ in terms of $t$? We can't read your mind to see what was on the page you were looking at when you had these thoughts. Show us. – David K Sep 05 '20 at 18:42
  • Sorry for not making it clear enough. I've changed the question, is it okay now? – BoostedAnimal Sep 05 '20 at 18:43
  • @BoostedAnimal No its not. Try using \int in latex braces to show us the $\int$ you are trying to evaluate. Less natural language, more math! – HelloWorld Sep 05 '20 at 18:44
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    Actually more natural language and more math. More of everything. – David K Sep 05 '20 at 18:47
  • @BoostedAnimal I've edited your MathJax and replaced the word "roots" with "radicals", hope this is what you had in mind. (I did not want to edit anything more substantially than that.) Having said that, what problem do you see with the substitution $t=\sqrt[10]{x}$ and $\sqrt{x}=t^5, \sqrt[5]{x^2}=t^4$ as suggested in the answers? –  Sep 05 '20 at 18:51
  • The problem is that the answer has a large curly brace that implies that $x=t^{10}$ can be derived from both roots and I just can't find how is that done. I've added the integral as well. I've also added an image. – BoostedAnimal Sep 05 '20 at 18:54
  • The answer key actually shows one of the ways that all symbols and no words can be a bad thing. The big curly brace between the equations with the radicals and the equation $x:=t^{10}$ doesn't actually mean anything mathematically--it's just a hint that the two things on the left, taken together, are somehow related to the thing on the right, and leaves you to guess what that relationship is. In this case, the relationship is not "therefore it follows logically that ...," the relationship is "so let's see what happens if we make the following substitution." – David K Sep 05 '20 at 22:10

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First of all you should review the definition of a "rational function." It is a function that can be written as a ratio of two polynomials.

The integrand $\frac{1}{x(\sqrt{x} + \sqrt[5]{x^2})}$ is not a rational function of $x.$ There is no way to make it equal to $\frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomial functions.

It is also not true that you "must" substitute $t = \sqrt[10]x.$ It is entirely possible that there is a completely different way to solve this integral.

But you book is showing you a way to solve this integral, and that way involves a $u$-substitution that leaves you with a rational function in the integrand. The trick to this particular substitution is to find something that lets you rewrite all the powers of $x$ as integer powers of your substituted variable, so that when you add and multiply as shown you'll have a polynomial in your new variable. (It's also necessary that whatever happens in the numerator due to the substitution, that also has to end up as a polynomial in your new variable. We don't know this will happen until we try it.)

The choice $t = \sqrt[{10}]{x}$ is by no means the only way to do this. But it does allow you to write each of the terms $x$, $\sqrt x$, and $\sqrt[10]x$ as an integer power of $t.$

So, if you did not have the answer key in front of you, how might you come to the decision to try $t = \sqrt[{10}]{x}$?

You could first convert everything into integer or fractional powers of $x$:

\begin{align} \sqrt x &= x^{1/2}, \\ \sqrt[5]{x^2} &= x^{2/5}, \\ \end{align}

so now we have $$ \frac{1}{x(x^{1/2} + x^{2/5})} .$$

Now for $x^{1/2}$ and $x^{2/5}$ both to be integer powers of the same number, it's sufficient for that number to be a power of $x$ with an exponent that divides both $\frac12$ and $\frac25$ evenly. How to find a number that divides two fractions evenly? Try putting them in terms of a common denominator,

$$ \frac12 = \frac{5}{10}, \quad \frac25 = \frac{4}{10}. $$

that is, $\frac12 = 5 \times \frac1{10},$ so $x^{1/2} = (x^{1/10})^5.$ And $\frac25 = 4 \times \frac1{10},$ so $x^{2/5} = (x^{1/10})^4.$

What is inside the parentheses in both cases? It's $x^{1/10}.$ So if we suppose that $t = x^{1/10},$ then $$ x^{1/2} + x^{2/5} = t^4 + t^5 . $$

Remember, this is not the only way to write these terms as integer powers of something. It just happens to be a convenient way to do it.

Again, there is often more than one way to solve an integral. A solution shows you how you can solve it, not how you must solve it. Especially with $u$-substitutions, unless you've seen the exact same problem before (possibly in a more general form), you're really just guessing that a particular substitution is going to help. You find out that it was a good one when you actually do it and solve the integral.

David K
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  • I still don't get how to get x i.e $t=^{10}\sqrt(x)$ from both of the roots. – BoostedAnimal Sep 05 '20 at 19:12
  • @BoostedAnimal You don't get $t=\sqrt[10]{x}$. You let it. David actually analyzed in great detail what is going on in the image you posted. Consider reading David's answer more carefully and awarding him by upvoting it or/and accepting it. – HelloWorld Sep 05 '20 at 20:35
  • The initial integral is $\int\frac{dx}{x(\sqrt{x} + \sqrt[5]{x^2})}$. When you substitute $t=\sqrt[10]{x}$, you must also find $dt$. Hence $t=\sqrt[10]{x} \iff \boxed{x = t^{10}} \iff dx = (t^{10})'dt \iff \boxed{dx = 10 t^9 dt}$. – HelloWorld Sep 05 '20 at 20:48
  • Hence, by substituting the boxed values into the initial integral you get $\int\frac{dx}{x(\sqrt{x} + \sqrt[5]{x^2})} = \int\frac{10 \cdot t^9 \cdot dt}{t^{10}(\sqrt{t^{10}} + \sqrt[5]{{t^{10}}^2})} = \int\frac{10 \cdot t^9 \cdot dt}{t^{10}(t^{\frac{10}{2}} + {t^{\frac{20}{2}}})} = \int\frac{10 \cdot dt}{t(t^{\frac{10}{5}} + {t^{\frac{20}{5}}})} = \int\frac{10 \cdot dt}{t(t^5 + {t^{4}})} = \boxed{ 10 \int\frac{ dt}{t^5(t + 1)}}$ @BoostedAnimal – HelloWorld Sep 05 '20 at 20:48