Where, exactly, did you get lost?

It should be clear that \(\displaystyle x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1\) and so \(\displaystyle \int_0^2 \sqrt{x^3+ 1}+ \sqrt[3]{x^2+ 2x} dx= \int_0^2 \sqrt{x^3+1}dx+ \int_0^2\sqrt[3]{(x+1)^2- 1}dx\).

Now, make that very clever substitution simplependulum gives: \(\displaystyle y^3= (x+1)^2- 1 \). Then \(\displaystyle 3y^2 dy= 2(x+1)dx\) and \(\displaystyle \sqrt[3]{(x+1)^2- 1}=\sqrt[3]{y^3}= y\).

Since \(\displaystyle y^3= (x+1)^2- 1\), \(\displaystyle (x+1)^2= y^3+ 1\) and \(\displaystyle x+ 1= \sqrt{y^3+ 1}\). That means that \(\displaystyle 3y^2dy= 2(x+1)dx= 2\sqrt{y^3+ 1} dx\) so that \(\displaystyle dx= \frac{3y^2}{2\sqrt{y^3+ 1}}dy= dx\).

Also, when x= 0, \(\displaystyle y^3= (1+1)^2- 1= 0\) and when x= 2, \(\displaystyle y^3= (1+2)^2- 1= 8\) so that y= 0 and 2 are still the limits of integration.

Putting that all together, \(\displaystyle \int_0^2 \sqrt[3]{(x+1)^2- 1}dx\)\(\displaystyle = \int_0^2\left(\sqrt[3]{y^3+ 1}\right)\left(\frac{3y^2}{2\sqrt{y^3+ 1}}dy\right)\)\(\displaystyle \frac{3}{2}\int_0^2\frac{y^3}{\sqrt{y^3+ 1}}dy\).

Now, that "y" is a dummy index- the actual integral is a number, not a function of y so we are free to replace it with anything we want. If just replace "y" with "x" (I think that's the "dummy substitution" simplependulum refers to. His first substitution is not "dummy" in any sense!) we get

\(\displaystyle \int_0^2 \frac{x^3}{\sqrt{x^3+ 1}}dx\).

Now put those two integrals back together:

\(\displaystyle \int_0^2 \sqrt{x^3+ 1}+ \frac{3}{2}\frac{x^3}{\sqrt{x^3+ 1}}dx\)

To "add fractions" get a "common denominator" by multiplying both numerator and denominator by \(\displaystyle 2\sqrt{x^3+ 1}\): \(\displaystyle \sqrt{x^3+1}\frac{2\sqrt{x^3+ 1}}{2\sqrt{x^3+ 1}}\)\(\displaystyle = \frac{2x^3+ 2}{2\sqrt{x^3+ 1}}\). (In effect, we are "rationalizing" the numerator.)

That gives \(\displaystyle \int_0^2 \frac{2x^3+ 2}{2\sqrt{x^3+ 1}}+ \frac{3x^3}{2\sqrt{x^3+ 1}}dx\)\(\displaystyle = \int_0^2 \frac{5x^3+ 2}{2\sqrt{x^3+ 1}}dx\).

simplependulum's next step was to multiply both numerator and denominator by x: \(\displaystyle \int_0^2 \frac{x(5x^3+ 2)}{2x\sqrt{x^3+ 1}}dx\). Of course, inside the square root that x becomes \(\displaystyle x^2\) so we have

\(\displaystyle \int_0^2 \frac{5x^4+ 2x}{2\sqrt{x^5+ x^2}}dx\).

And simplependulum did that because then the derivative of \(\displaystyle x^5+ x^2\) is precisely \(\displaystyle 5x^4+ 2x\)!

The substitution \(\displaystyle u= x^5+ x^2\) gives \(\displaystyle du= (5x^4+ 2x)dx\) so that \(\displaystyle \int_0^2\frac{5x^4+ 2x}{2\sqrt{x^5+ x^2}}dx\)\(\displaystyle = \int_0^36 \frac{du}{2\sqrt{u}}= \frac{1}{2}\int_0^36 u^{-1/2}du\).

That's my explanation. I don't pretend that I would have found that myself!