Integrali, metodi a confronto.

$ \int x \sqrt[3]{x-2} \, dx = \, ⊳ $

a.   Come suggerito, poniamo $ t = x-2 \; ⇒ \; x = t+2 \; ⇒ \; dx = dt $

$ ⊳ \, = \int(t+2)t^{\frac{1}{3}} \, dt = \int t^{\frac{4}{3}} \, dt + 2 \int t^{\frac{1}{3}} \, dt = \frac{3}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}} + c = t^{\frac{4}{3}} \left(\frac{3}{7}t + \frac{3}{2} \right) + c = (x-2) \cdot \sqrt[3]{x-2} \left(\frac{6x}{14}- \frac{12}{14} + \frac{21}{14} \right) + c = \frac{3}{14}(2x+3)\sqrt[3]{x-2} + c $ 

b.  Come osservato. $ x \sqrt[3]{x-2} = (x-2+2) \sqrt[3]{x-2} = (x-2)\sqrt[3]{x-2} + 2 \sqrt[3]{x-2} = \sqrt[3]{(x-2)^4} + 2 \sqrt[3]{x-2} $

Integrando

$ ⊳ \, = \int \sqrt[3]{(x-2)^4} \, dx + \int \sqrt[3]{x-2} \, dx =  \int (x-2)^{\frac{4}{3}} \, dx + \int (x-2)^{\frac{1}{3}} \, dx = \frac{3}{7}(x-2)^{\frac{7}{3}} + \frac{3}{2}(x-2)^{\frac{4}{3}} + c = (x-2)^{\frac{4}{3}} \left( \frac{3}{7} x + \frac{3}{2} \right) = \frac{3}{14}(x-2)(2x+3)\sqrt[3]{x-2} + c $