Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$? 
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?

I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \frac{t^2 - 1}{p}t \,dt = \frac{1}{p}\int_1^\sqrt{p+1} t^3 - t \,dt = \frac{1}{p}\bigg[\frac{t^4}{4}\bigg\vert_1^\sqrt{p+1} - \frac{t^2}{2}\bigg\vert_1^\sqrt{p+1}\bigg]$$
$$ I =  \frac{1}{p} \bigg[\frac{1}{4}(p^2 + 2p + 1 - 1) - \frac{1}{2}(p + 1 - 1) \bigg] = \frac{1}{p}\frac{1}{4}p^2 = \frac{1}{4}p $$
So for $p=4$, the integral should have the value $1$, but WA doesn't agree: http://wolfr.am/5p-i9uOA
I looked at it for quite a while but I can't locate the mistake. Since $p>0$, everything seems fine to me but apparently it isn't.

$$I(p)=\int_{0}^{1}x\sqrt{px+1}\,dx = \frac{1}{p^2}\int_{0}^{p}x\sqrt{x+1}\,dx $$
but integration by parts gives:
$$ \int x\sqrt{x+1}\,dx = \frac{2}{15}(3x-2)(1+x)^{3/2} $$
hence:
$$ I(p) = \frac{4}{15p^2}+\frac{2}{15}\sqrt{1+p}\left(3+\frac{1}{p}-\frac{2}{p^2}\right).$$
