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 $I=\frac{4}{15}(x+1)^{5/2}$.
 A: 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.
A: It's easier to do this using the substitution $u=x+1$. Then
$$I=\int(u-1)u^{\frac12}du=\frac25u^{5/2}-\frac23u^{3/2}+C,$$
where $C$ is the constant of integration.
Substitute back for $x$ in this to get
$$
I=(1+x)^{3/2}\left(\frac25(1+x)-\frac23\right)+C=(1+x)^{3/2}\left(\frac25 x - {4\over 15}\right)+C
$$
No matter how you wangle it, there isn't a term of the form $\frac23(1+x)^{5/2}$ here, so whereever you got that from is incorrect.
A: Let $u=\sqrt{x+1}$ and then
$$ x=u^2-1, dx=2udu. $$
so
$$I=\int x(x+1)^{1/2} dx=2\int(u^2-1)u^2du=\frac25u^5-\frac23u^3+C=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C.$$
