For calculus lovers

$\displaystyle{f(x)~ =~ \sqrt [ 3 ]{ 1+x^{ 2 }} }$

$\displaystyle{ f(-x)~ =~ \sqrt [ 3 ]{ 1+x^{ 2 }} }$

$\displaystyle{ f(x)~ =~ f(-x)~ \cdots \cdots \cdots (therefore~ f(x)~ is~ even~ function)}$

$\displaystyle{ g(t)~ =~ \int _{ 0 }^{ t }{ f(x)dx ~ \cdots \cdots \cdots \cdots (therefore~ g(t)~ is~ odd\quad function) }}$

$\displaystyle{let~ \int _{ ln\left( tan\left( \frac { \pi }{ 6 } \right) \right) }^{ ln\left( tan\left( \frac { \pi }{ 3 } \right) \right) }{ g(x)dx~ =~ I }}$

$\displaystyle{\therefore ~ I~ =~ \int _{ -ln\left( \sqrt { 3 } \right) }^{ ln\left( \sqrt { 3 } \right) }{ g(x)dx }}$

$\displaystyle{\because ~ \int _{ -a }^{ a }{ f(x)dx~ =~ 0~ \cdots \cdots \cdots (where~ f(x)~ is~ a~ odd~ funtion) } }$

$\therefore I~ =~ 0$