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?
 A: 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 $$\begin{align}\int t\,\sqrt{1+t}\;dt & = \frac 23 t\,(1+t)^{3/2} - \frac 23 \int (1+t)^{3/2}\, dt\\ \\ & = \frac 23 t\,(1+t)^{3/2} - \frac 4{15}(1+t)^{5/2} + C\end{align}$$
A: $$\int \underbrace{t}_{=u}\underbrace{\sqrt{1+t}}_{=v'}dt=\frac{2}3t(1+t)^{3/2}-\frac23\int(1+t)^{3/2}dt=\frac{2}3t(1+t)^{3/2}-\frac4{15}(1+t)^{5/2}+C$$
A: This is a case where the LIATE rule, when applying the integration by parts technique, actually helps. So we differentiate the factor $t$ and integrate the factor $\left(
1+t\right) ^{1/2}$ of the product $t\cdot \left( 1+t\right) ^{1/2}$ in the integrand. Since $$\int \left( 1+t\right) ^{1/2}\,dt=\frac{2}{3} \left(
1+t\right) ^{3/2}\,dt,$$
we have that
$$I =\int t\cdot \left( 1+t\right) ^{1/2}dt=t\cdot \frac{2}{3}\left( 1+t\right) ^{3/2}-\int 1\cdot\frac{2}{3} \left(
1+t\right) ^{3/2}\,dt. $$
A: Just another approach without "integration by part" method.
Let $u^2=1+t$ so you have $dt=2udu$ and $t=u^2-1$.
\begin{align}
\int t \sqrt{t+1} dt &= 2\int(u^2-1)u^2 du \\
&= 2 \int (u^4-u^2)du\\
&= 2\left(\frac15 u^5 -\frac13 u^3\right)+C\\
&= 2\left(\frac15 (1+t)^2\sqrt{1+t} -\frac13 (1+t)\sqrt{1+t}\right)+C\\
&=\frac{2(1+t)(3t-2)}{15}\sqrt{1+t}+C
\end{align}
A: $$\int t \sqrt(1+t)=\frac{2}{3}t(1+t)^{3/2}-2/3\int(1+t)^{3/2}dt$$
$$=\frac{2}{3}t(1+t)^{3/2}-4/15(1+t)^{5/2}+Q$$
