If $g(x)=∫\frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{6}}}$, then $g(1) - g(0)$ is:

$\frac{3π}{2}-4$

The correct answer is Option (4) → $\frac{3π}{2}-4$

$g(x)=∫\frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{6}}}$

$=∫\frac{dx}{x^{\frac{1}{6}}(x^{\frac{1}{3}}+1)}$

let $t=x^{\frac{1}{3}}$

$⇒x=t^3$ and $x^{\frac{1}{6}}=t^{\frac{1}{2}}$

Now,

$dx=3t^2dt$

Substituting this,

$g(x)=\int\frac{3t^2dt}{t^{\frac{1}{2}}(t+1)}dt$

$=3\int\left(t^{\frac{1}{2}}-\frac{t^{\frac{1}{2}}}{t+1}\right)dt$

$=2t^{\frac{3}{2}}-3.\left(2\sqrt{t}-2\tan^{-1}(\sqrt{t})\right)+C$

$=2x^{\frac{1}{2}}-6x^{\frac{1}{6}}+6\tan^{-1}(x^{\frac{1}{6}})+C$

$∴g(1)-g(0)=\left[-4+\frac{3π}{2}+C-0\right]=\frac{3π}{2}-4$