Problem:
$$\displaystyle \int \frac{\sqrt[3]{x}}{\sqrt[3]{x^2} -\sqrt{x} } dx$$
My attempts:
$$\displaystyle \int \frac{\sqrt[3]{x}}{\sqrt[3]{x^2} -\sqrt{x} } dx={\displaystyle\int} \dfrac{1}{6x^\frac{5}{6}}\cdot \dfrac{6x^\frac{2}{3}}{\sqrt[6]{x}-1}\,\mathrm{d}x$$
Substitute $u=\sqrt[6]{x} \longrightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{6x^\frac{5}{6}} \longrightarrow \mathrm{d}x=6x^\frac{5}{6}\,\mathrm{d}u$, we have
$$\sqrt{x}=u^3$$
$$x^\frac{2}{3}=u^4$$
$$x^\frac{7}{6}=u^7$$
$$=\displaystyle\int \dfrac{u^4}{u-1}\,\mathrm{d}u$$
Now, let's solving
$${\displaystyle\int}\dfrac{u^4}{u-1}\,\mathrm{d}u$$
Substitute $v=u-1 \longrightarrow \dfrac{\mathrm{d}v}{\mathrm{d}u} = 1 \longrightarrow \mathrm{d}u=\mathrm{d}v$, we have
$$u^4=\left(v+1\right)^4$$
$$={\displaystyle\int}\dfrac{\left(v+1\right)^4}{v}\,\mathrm{d}v={\displaystyle\int}\left(v^3+4v^2+6v+\dfrac{1}{v}+4\right)\mathrm{d}v={\displaystyle\int}v^3\,\mathrm{d}v+4{\displaystyle\int}v^2\,\mathrm{d}v+6{\displaystyle\int}v\,\mathrm{d}v+{\displaystyle\int}\dfrac{1}{v}\,\mathrm{d}v+4{\displaystyle\int}1\,\mathrm{d}v=\ln\left(|v|\right)+\dfrac{v^4}{4}+\dfrac{4v^3}{3}+3v^2+4v=4\left(u-1\right)+\dfrac{\left(u-1\right)^4}{4}+\dfrac{4\left(u-1\right)^3}{3}+3\left(u-1\right)^2+\ln\left(|u-1|\right)$$
Keeping all this in mind, we get
$$\displaystyle \int \frac{\sqrt[3]{x} }{\sqrt[3]{x^2} -\sqrt{x} } dx=\displaystyle\int\dfrac{u^4}{u-1}\,\mathrm{d}u =24\left(u-1\right)+\dfrac{3\left(u-1\right)^4}{2}+8\left(u-1\right)^3+18\left(u-1\right)^2+6\ln\left(|u-1|\right)+C=24\left(\sqrt[6]{x}-1\right)+\dfrac{3\left(\sqrt[6]{x}-1\right)^4}{2}+8\left(\sqrt[6]{x}-1\right)^3+18\left(\sqrt[6]{x}-1\right)^2+6\ln\left(\left|\sqrt[6]{x}-1\right|\right)+C=\dfrac{3x^\frac{2}{3}}{2}+2\sqrt{x}+3\sqrt[3]{x}+6\left(\sqrt[6]{x}+\ln\left(\left|\sqrt[6]{x}-1\right|\right)\right)+C=6\left(\dfrac{3x^\frac{2}{3}+4\sqrt{x}+6\sqrt[3]{x}+12\sqrt[6]{x}}{12}+\ln\left(\left|\sqrt[6]{x}-1\right|\right)\right)+C$$
I tried to solve this question in WA before sending it here.
Wolfram Alpha gave me :
Because, before I solved the integral myself, I had to find out if my answer contained elementary functions or not. This is because I learn integrals that can only be solved with basic functions. I only know the "name" of special functions. Then I realized it was an "input error." But unfortunately it wasn't my fault. I didn't write an entry using the math syntax. But my inputs work perfectly even with integrals that cannot be solved by elementary functions. Here are some examples:
$$\displaystyle \int \sin(x^2)dx$$
An integral containing generalized hypergeometric function, error and imaginary error functions: $$\displaystyle \int \sin(x^2) \log(x)dx$$
I can give more complicated integrals, which works my standart input. I reported the situation to Wolfram Alpha. They acknowledged that this was "their fault." And they said they'd fix it as soon as possible. The reason I wrote this is because of some humiliating sarcasm that some high reputation users do to me. Maybe I was supposed to write this error in another category. I got some roughly comments for the question. I had to show that it wasn't my "invalid input" . Thank you very much.
$$ \int \frac{\sqrt[3]{x} }{\sqrt[3]{x^2} -\sqrt{x} } dx = 6\int \frac{\sqrt[3]{t^6} }{\sqrt[3]{t^{12}} -\sqrt{t^6} }t^5\,dt = 6\int \frac{t^4}{t-1}\,dt= 6\int\left(\frac{t^4-1}{t-1}+\frac{1}{t-1}\right)\,dt.$$
How could this have no closed-form expression ?