an upper bound for number of primes in the interval $[n^2+n,n^2+2n]$ What is an upper bound for the number of primes in an interval of $n$ consecutive 
numbers? 
What is an upper bound for the number of primes in the interval $[n^2+n,n^2+2n]$?

According to the Prime-counting function
$$\text{number of primes below natural }n = \pi(n) \approx \frac{n}{\ln n}$$
so you just want an approximation of $\pi(n^2+2n) - \pi(n^2+n)$:
$$\begin{align}
\pi(n^2+2n) - \pi(n^2+n) &\approx \frac{n^2+2n}{\ln (n^2+2n)} - \frac{n^2+n}{\ln (n^2+n)}\\
&\approx \lim_{n \to \infty} \left( \frac{n^2+2n}{\ln (n^2+2n)} - \frac{n^2+n}{\ln (n^2+n)} \right)\\
&= \frac{n \left(-2 \ln \left(\frac{1}{n}\right)-1\right)}{4 \ln ^2\left(\frac{1}{n}\right)} - \frac{3 \left(\ln \left(\frac{1}{n}\right)+1\right)}{8 \ln ^3\left(\frac{1}{n}\right)}+O\left(\frac{1}{n}\right)\\
& \approx \frac{n \left(2 \ln n-1\right)}{4 \ln ^2n} - \frac{3 \left(-\ln n+1\right)}{-8 \ln ^3n}\\
& = \frac{n \left(2 \ln n-1\right)}{4 \ln ^2n} - \frac{3 \left(\ln n-1\right)}{8 \ln ^3n}
\end{align}$$
For example, let $n = 1000$, now you want to calculate the number of primes $\in [1000\cdot 1001, 1000\cdot 1002] \equiv [1001000, 1002000]$
and that's approximately
$$\frac{1000 \left(2 \ln 1000-1\right)}{4 \ln ^21000} - \frac{3 \left(\ln 1000-1\right)}{8 \ln ^31000} = 67.1365$$
while the real number is
$$78649 - 78572 = 77$$
so that's an error of about $13$% with $n = 1000$. Note that you'll be able to obtain better approximations either developing the series at $\infty$ to greater orders of magnitude, either taking $n$ very large.
