Integrali, metodi a confronto.

a.  Poniamo $ t = x+1 \; ⇒ \; x = t - 1 \; ⇒ \; dx = dt $

$ \int \frac {x}{\sqrt {x+1}} \, dx = \int \frac {t-1}{t^{\frac{1}{2}}} \, dt =  \int t^{\frac{1}{2}} - t^{-\frac{1}{2}} \, dt = \frac{2}{3}t^{\frac{3}{2}} - 2 t^{\frac{1}{2}} \, dt  + c = t^{\frac{1}{2}} \left(\frac{2}{3}t - 2 \right) + c = (x+1)^{\frac{1}{2}} \left(\frac{2}{3}(x+1)- 2 \right) + c = \frac{2}{3}\sqrt{x+1}(x+1-3) + c = \frac{2}{3}(x-2) \sqrt{x+1} + c $

b.  $ \frac {x}{\sqrt{x+1}} = \frac {x+1 -1}{\sqrt{x+1}} = \sqrt{x+1} - \frac{1}{\sqrt{x+1}} = (x+1)^{\frac{1}{2}}  -  (x+1)^{-\frac{1}{2}} $

Integrando

$ \int \frac {x}{\sqrt {x+1}} \, dx = \int(x+1)^{\frac{1}{2}} \, dx - \int (x+1)^{-\frac{1}{2}} \, dx = \frac{2}{3}(x+1)^{\frac{3}{2}} - 2 (x+1)^{\frac{1}{2}} + c = \frac{2}{3}(x+1)^{\frac{1}{2}} (x+1-3) +c  = \frac{2}{3}(x-2) \sqrt{x+1} + c $