2
$\begingroup$

I thought that the Substitution rule required for the part equal to $du$ to be in the integral somewhere. However I came across this one and am not entirely sure why the solution is that way. Link to Integral Calculator

$$ \int \frac{1}{\sqrt[3]{x} + x} dx $$

The calculator uses substitution

$$ u = \sqrt[3]{x} $$ $$ du = (\sqrt[3]{x})'dx = \frac{1}{3x^{2/3}} \Rightarrow \frac{dx}{x^{3/2}} = 3du $$

Now if we extract $x^{2/3}$ from the denominator it makes sense to me and we arrive at

$$ \int \frac{1}{x^{2/3}(x + x^{1/3})}dx $$

However the integral calculator i linked does something else

$$ \int \frac{1}{\sqrt[3]{x} + x}dx = \int \frac{3u^2}{u^3 + u}du $$

and then it continues with the solution. My missing point is the $ 3u^2 $ part. Where did that come from ? I understand that its the derivative of $ u^3 $ but how did we just plug it in there? I am probably missing something super obvious, but was not able to wrap my head around it.

$\endgroup$
3
  • 1
    $\begingroup$ In your work, you have the even more complicated example of having something on the $\mathrm dx$ side and something on the $\mathrm du$ side ;-) $\endgroup$ Commented Jul 14, 2019 at 22:00
  • $\begingroup$ without having the possibility to just divide? $\endgroup$ Commented Jul 16, 2019 at 15:57
  • 1
    $\begingroup$ Since it is a constant, you could divide it over, yes. But if you had something like $\frac{\mathrm dx}{x^{3/2}}=u~\mathrm du$, then no, you wouldn't. $\endgroup$ Commented Jul 16, 2019 at 17:05

3 Answers 3

Reset to default
10
$\begingroup$

You've substituted $\color{red}{u = x^{1/3}} \implies du = \frac{1}{3}x^{-2/3}dx \implies dx = 3x^{2/3}du \implies \color{red}{dx = 3u^2du}$

Also, $\color{red}{x = u^3}$

Hence $$\int\frac{\color{blue}{dx}}{x^{1/3}+x} = \int\frac{\color{blue}{3u^2du}}{u+u^3}$$

$\endgroup$
5
$\begingroup$

The old way I learnt.

$$u = \sqrt[3]{x}\implies x=u^3\implies dx=3u^2 du$$ $$\int \frac{dx}{\sqrt[3]{x} + x} =3\int \frac{u^2}{u+u^3 }\,du=3\int \frac{u}{1+u^2 }\,du$$

$\endgroup$
2
$\begingroup$

From the equation \begin{align} du = \dfrac{dx}{3x^{2/3}}, \end{align} we get \begin{align} dx = 3x^{2/3} \cdot du = 3u^2\, du \end{align} (because $x^{1/3} = u$) Hence, \begin{align} \int \dfrac{1}{x^{1/3} + x} \, dx &= \int \dfrac{1}{u + u^3} (3u^2 \, du) \\ &= \int \dfrac{3u^2}{u^3 + u} \, du \end{align}

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.