int(dx)/(root3xroot3((x+5)^5))=? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int \frac{dx}{\sqrt[3]{x}\sqrt[3]{(x+5)^5}}\) \(I = \int \frac{dx}{\sqrt[3]{x(x+5)^5}}\) \(I = \int \frac{dx}{\sqrt[3]{x \cdot x^5(1+\frac{5}{x})^5}}\) \(I = \int \frac{dx}{\sqrt[3]{x^6(1+\frac{5}{x})^5}}\) \(I = \int \frac{dx}{x^2\sqrt[3]{(1+\frac{5}{x})^5}}\) \(I = \int \frac{dx}{x^2(1+\frac{5}{x})^{\frac{5}{3}}}\) এখন, ধরি \(u = 1 + \frac{5}{x}\) তাহলে, \(\frac{du}{dx} = -\frac{5}{x^2}\) সুতরাং, \(dx = -\frac{x^2}{5}du\) এখন, \(I = \int \frac{1}{x^2 u^{\frac{5}{3}}} \cdot (-\frac{x^2}{5})du\) \(I = -\frac{1}{5} \int \frac{1}{u^{\frac{5}{3}}}du\) \(I = -\frac{1}{5} \int u^{-\frac{5}{3}}du\) \(I = -\frac{1}{5} \cdot \frac{u^{-\frac{5}{3}+1}}{-\frac{5}{3}+1} + c\) \(I = -\frac{1}{5} \cdot \frac{u^{-\frac{2}{3}}}{-\frac{2}{3}} + c\) \(I = \frac{1}{5} \cdot \frac{3}{2} u^{-\frac{2}{3}} + c\) \(I = \frac{3}{10} u^{-\frac{2}{3}} + c\) \(I = \frac{3}{10} (1+\frac{5}{x})^{-\frac{2}{3}} + c\) অতএব, \(\int \frac{dx}{\sqrt[3]{x}\sqrt[3]{(x+5)^5}} = \frac{3}{10}(1+\frac{5}{x})^{-\frac{2}{3}} + c\) 🎉