$\int \frac{d x}{\sqrt[3]{1-6 x}}=f(x)+c$ হলে $f(x)=$ কত ?

$\frac{1}{4}(1-6 x)^{2 / 3}$

$\frac{1}{4}(1-6 x)^{3 / 2}$

$-\frac{1}{4}(1-6 x)^{2 / 3}$

$-\frac{1}{4}(1-6 x)^{3 / 2}$