<s>[INST] Calculate $\int x(x+1)^{1/2} dx$. The integral in question:

$$I=\int x(x+1)^{1/2} dx$$

I've done integration by parts to get to
$I=x\frac{2}{3}(x+1)^{3/2} - \int\frac{2}{3}(x+1)^{3/2} dx$.
I've used mathematica and for the integral $\int \frac{2}{3}(x+1)^{3/2} dx$ when I calculate it by hand, I get $\frac{4}{15}(x+1)^{5/2}$ but apparently it should be just $\frac{2}{3}(x+1)^{5/2}$. I think something is wrong with my arithmetic. I thought $\int x^ndx = x^{n+1}/(n+1)$ which, in my case, I thought it would be $n+1=\frac{5}{2}$, $\frac{2}{3}\colon\frac{5}{2} = 4/15$. Therefore $I=\frac{4}{15}(x+1)^{5/2}$.
 [/INST] I use $$\begin{align} x\sqrt{1+x}&=(1+x)\sqrt{1+x}-\sqrt{1+x}\\&=(1+x)^{3/2}-(1+x)^{1/2}.\end{align} $$ With the right side easily integrated to give (ignoring the constant:)
$$\begin{align} \frac25(1+x)^{5/2}-\frac23(1+x)^{3/2}&=\left(\frac25(1+x)-\frac23\right)(1+x)^{3/2}\\&=\left(\frac25x-\frac4{15}\right)(1+x)^{3/2}\end{align} $$
Your answer is: $$\begin{align} \frac23x(1+x)^{3/2}-\frac4{15}(1+x)^{5/2}&=\left(\frac23x-\frac4{15}(1+x)\right)(1+x)^{3/2}\\&=\left(\frac25x-\frac4{15}\right)(1+x)^{3/2} \end{align} $$
Same answer.
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