Integral clarification

Question

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.

Answer 1

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

Answer 2

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}

Answer 3

The old way I learnt.

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