মান নির্ণয় কর  ∫ 1/root(3)(1-6x) 

Explanation:

Another Explanation (5): সমাধান: ধরি, \( I = \int \frac{1}{\sqrt[3]{1-6x}} dx \) এখন, \( 1-6x = u \) ধরি। 🤔 তাহলে, \( -6 dx = du \) বা, \( dx = -\frac{1}{6} du \) সুতরাং, \( I = \int \frac{1}{\sqrt[3]{u}} \left(-\frac{1}{6}\right) du \) \( = -\frac{1}{6} \int u^{-\frac{1}{3}} du \) আমরা জানি, \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \) অতএব, \( I = -\frac{1}{6} \cdot \frac{u^{-\frac{1}{3}+1}}{-\frac{1}{3}+1} + C \) \( = -\frac{1}{6} \cdot \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + C \) \( = -\frac{1}{6} \cdot \frac{3}{2} u^{\frac{2}{3}} + C \) \( = -\frac{1}{4} u^{\frac{2}{3}} + C \) যেহেতু \( u = 1-6x \), তাই \( I = -\frac{1}{4} (1-6x)^{\frac{2}{3}} + C \) সুতরাং, \( \int \frac{1}{\sqrt[3]{1-6x}} dx = -\frac{1}{4} (1-6x)^{\frac{2}{3}} + C \) 🎉 উত্তর: \( -\frac{1}{4}(1-6x)^{\frac{2}{3}} \)