$$ \int \frac{x^{-7/6}-x^{5/6}}{\sqrt[3]{x}\sqrt{x^2+x+1}-\sqrt{x}\sqrt[3]{x^2+x+1}}\, \mathrm{d}x $$
Its a homework question. Trigonometric substitutions make it very complex. I cannot think of any suitable method. Hints needed!
EDIT
Multiplying and dividing by $$x^{7/6}$$ gives :
$$ \displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\class{placeholder}{}}}_{\cssId{lower-bound-mathjax}{\class{placeholder}{}}} -\dfrac{1-\frac{1}{x^2}}{\sqrt{x+\frac{1}{x}+1}-\sqrt[3]{x+\frac{1}{x}+1}}\,\cssId{int-var-mathjax}{\mathrm{d}x} $$
then its just substitution $$ x+{1/x} = t $$
$$I=\int \frac{x^{-7/6}-x^{5/6}}{\sqrt[3]{x}\sqrt{x^2+x+1}-\sqrt{x}\sqrt[3]{x^2+x+1}}\, \mathrm{d}x$$ $$I=\int \frac{x^{-7/6}(1-x^2)}{\sqrt[3]{x}\sqrt{x}\sqrt{(x+\frac1x)+1}-\sqrt{x}\sqrt[3]{x}\sqrt[3]{(x+\frac1x)+1}}\, \mathrm{d}x$$ $$I=\int -\frac{1-\frac{1}{x^2}}{\sqrt{(x+\frac1x)+1}-\sqrt[3]{(x+\frac1x)+1}}\, \mathrm{d}x$$ now let $x+\dfrac1x=u^6$ then $$I=\int-\dfrac{6u^5}{u^3-u^2}\ du$$