int1/(root(3)(1-6x))dx=? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \( u = 1 - 6x \). তাহলে, \( du = -6 dx \) অথবা \( dx = -\frac{1}{6} du \). অতএব, \[ \int \frac{1}{\sqrt[3]{1-6x}} dx = \int \frac{1}{\sqrt[3]{u}} \left(-\frac{1}{6}\right) du = -\frac{1}{6} \int u^{-\frac{1}{3}} du \] এখন, ইন্টিগ্রেশন করি: \[ -\frac{1}{6} \int u^{-\frac{1}{3}} du = -\frac{1}{6} \cdot \frac{u^{-\frac{1}{3} + 1}}{-\frac{1}{3} + 1} + C = -\frac{1}{6} \cdot \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + C = -\frac{1}{6} \cdot \frac{3}{2} u^{\frac{2}{3}} + C = -\frac{1}{4} u^{\frac{2}{3}} + C \] এখন, \( u \) এর মান বসিয়ে পাই: \[ -\frac{1}{4} (1-6x)^{\frac{2}{3}} + C \] সুতরাং, \[ \int \frac{1}{\sqrt[3]{1-6x}} dx = -\frac{1}{4} (1-6x)^{\frac{2}{3}} + C \] এখানে \(C\) হলো integration constant. সুতরাং নির্ণেয় সমাধান: \(-\frac{1}{4}(1-6x)^{\frac{2}{3}}\) 🎉