Integral with un-intuitive U-Substitution On a recent midterm we were given the following integral and were expected to integrate it.
$$\int_{-27}^{-8} \frac{\mathrm dx}{x + x^{2/3}}$$
 A: Coming to it cold early in my study of integration methods, I’d look at that denominator and think that letting $u=x^{1/3}$ would make it look a lot nicer: $u^3+u^2$. It would also give me nice limits of integration, from $u=-3$ to $u=-2$; on an exam problem that’s likely to be a good sign. Of course, the question then is what would happen to $dx$: would it make matters harder or easier? If $u=x^{1/3}$, then $u^3=x$, so $dx=3u^2\,du$, and we have
$$\int_{-3}^{-2}\frac{3u^2}{u^3+u^2}du=3\int_{-3}^{-2}\frac{du}{u+1}\;.$$
That’s a nice integral that you can do directly or, if necessary, by making the further substitution $v=u+1$, $dv=du$.
A: $$\int \frac{1}{x+x^{2/3}}=\int \frac{1}{x^{2/3}(x^{1/3}+1)}$$
Now substitute $u=x^{1/3}+1$
This removes the $x^{2/3}$ term and changes it into a well known integral, $$3\int \frac 1u=3\log|u|$$
Of course we have integration bounds to worry about so the substitution changes the bounds to $-1$ and $-2$. Therefore the answer would be $3\log|-1|-3\log|-2|=3\log 1-3\log 2=\boxed{-3\log 2}$
A: what happens if we change the variable $x^{1/3} = u, u^3 = x.$
$$\int_{-27}^{-8} \dfrac{dx}{x + x^{2/3}} = 
\int_{-3}^{-2}\dfrac{3u^2\,du}{u^3+u^2} = \dfrac{3\,du}{1+u} = 3\ln|1+u|_{-3}^{-2} = -3\ln 2$$
A: $$x+x^{\frac{2}{3}}=\\x^{\frac{3}{3}}+x^{\frac{2}{3}}=\\x^{\frac{2}{3}}(x^{\frac{1}{3}}+1)=\\\sqrt[3]{x^2}(\sqrt[3]{x}+1)$$now apply this$$u=\sqrt[3]{x}\\x=u^3\\dx=3u^2du\\\frac{1}{x+x^{\frac{2}{3}}}dx=\\\frac{1}{u^2(u+1)}*(3u^2du)=\\\frac{3du}{u+1}\\$$so $$\int\frac{1}{x+x^{\frac{2}{3}}}dx=\int  \frac{3du}{u+1}=3ln|u+1|+c\\=3ln|\sqrt[3]{x}+1|+c$$
