Integral by parts of this function I have this integral to solve:

$\int t \sqrt{1+t}\,dt$

The book says to solve it by part but i have no clue about how to do so. Can someone help me?
Apply the integration by parts formula, $$\int u\,\frac{dv}{dt}dt = uv - \int v\,\frac{du}{dt}dt$$
Start by letting $u = t$. Then $du = dt$, and $$dv = \sqrt{1+t} \;dt = (1+t)^{1/2} \;dt\implies v = \dfrac 23 (1+ t)^{3/2}$$

In this case, we have $$\begin{align}\int t\,\sqrt{1+t}\;dt &amp; = \frac 23 t\,(1+t)^{3/2} - \frac 23 \int (1+t)^{3/2}\, dt\\ \\ &amp; = \frac 23 t\,(1+t)^{3/2} - \frac 4{15}(1+t)^{5/2} + C\end{align}$$
$$\int t \sqrt(1+t)=\frac{2}{3}t(1+t)^{3/2}-2/3\int(1+t)^{3/2}dt$$
$$=\frac{2}{3}t(1+t)^{3/2}-4/15(1+t)^{5/2}+Q$$