মান নির্ণয় কর \int \frac{1}{\sqrt[3]{1-6 x}} d x \quad\left[=\int(1-6 x)^{-\frac{1}{3}} d x=\frac{(1-6 x)^{-\frac{1}{3}+1}}{\left(-\frac{1}{3}+1\right)(-6)}=\frac{(1-6 x)^{\frac{2}{3}}}{-4}\right]