int(dx)/(root3(1-6x))   এর যোজিত ফল -- 

Explanation:

Another Explanation (5): সমাধান: ধরি, \( u = 1 - 6x \) তাহলে, \( \frac{du}{dx} = -6 \) সুতরাং, \( dx = -\frac{1}{6} du \) এখন, \( \int \frac{dx}{\sqrt[3]{1-6x}} = \int \frac{-\frac{1}{6} du}{\sqrt[3]{u}} \) \( = -\frac{1}{6} \int u^{-\frac{1}{3}} du \) \( = -\frac{1}{6} \cdot \frac{u^{-\frac{1}{3} + 1}}{-\frac{1}{3} + 1} + C \) \( = -\frac{1}{6} \cdot \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + C \) \( = -\frac{1}{6} \cdot \frac{3}{2} u^{\frac{2}{3}} + C \) \( = -\frac{1}{4} u^{\frac{2}{3}} + C \) \( = -\frac{1}{4} (1-6x)^{\frac{2}{3}} + C \) অতএব, \( \int \frac{dx}{\sqrt[3]{1-6x}} = -\frac{1}{4} (1-6x)^{\frac{2}{3}} + C \) 🎉🎉 সুতরাং নির্ণেয় যোগফল \( -\frac{1}{4} (1-6x)^{\frac{2}{3}} \) 🥳