Substituting $x\rightarrow \frac{x}{u}=z$ and taking $1/u=y$ we define a function $f(y)=\frac{y}{(1+y)^{r/2}}-\int_{1}^{1+y} z^{-r} dz$ where $r=-p, t=t/2$.
Now, $f(0)=0$ and $f'(y)=\frac{(1+y)^t-ty(1+y)^{t-1}-1}{(1+y)^{t}}$.
Again, as $(1+y)>1$ we define $f_1(y)=(1+y)^{t+1}-ty(1+y)^t-(1+y), f_1(0)=0$.
Hence, $f'_1(y)=(1+y)^t-t^2y(1+y)^{t-1}-1$. So, we carry on doing this and get $f_n(y)=(1+y)^t-t^{n}y(1+y)^{t-1}-1 ;f_n(0)=0$.
For, $n>>0, t^n ≈0$ as $0\leq t \leq 1/2$ $f_n(y)=(1+y)^t-1>0 \rightarrow f_{n-1}(y)>0 \rightarrow ... \rightarrow f(y)>0$.
Hence, proved. So, it must hold for $-2\leq p \leq 0$