Answer: (a) \(\displaystyle \int \frac{x}{\sqrt{x^{2}-1}}\,dx=\sqrt{x^{2}-1}+C\).
(b) \(\displaystyle \sum_{r=2}^{N} \ln\left(r+\sqrt{r^{2}-1}\right)\gt N\ln\left(N+\sqrt{N^{2}-1}\right)-\sqrt{N^{2}-1}.\)
(c) A valid upper bound is
\(\displaystyle \sum_{r=2}^{N} \ln\left(r+\sqrt{r^{2}-1}\right)\lt (N+1)\ln\left(N+\sqrt{N^{2}-1}\right)-\sqrt{N^{2}-1}.\)
An alternative upper bound is
\(\displaystyle \sum_{r=2}^{N} \ln\left(r+\sqrt{r^{2}-1}\right)\lt (N+1)\ln\left(N+1+\sqrt{N^{2}+2N}\right)-\sqrt{N^{2}+2N}.\)
(a) Let \(u=x^{2}-1\). Then \(du=2x\,dx\), so \(x\,dx=\tfrac12\,du\).
Hence
\(\displaystyle \int \frac{x}{\sqrt{x^{2}-1}}\,dx=\frac12\int \frac{1}{\sqrt{u}}\,du=\frac12\int u^{-1/2}\,du=\sqrt{u}+C=\sqrt{x^{2}-1}+C.\)
(b) Since \(y=\cosh^{-1}x\) is increasing for \(x\ge 1\), the total area of the right-endpoint rectangles of width 1 from \(x=1\) to \(x=N\) is greater than the area under the curve:
\(\displaystyle \cosh^{-1}2+\cosh^{-1}3+\cdots+\cosh^{-1}N\gt \int_{1}^{N} \cosh^{-1}x\,dx.\)
Using \(\cosh^{-1}r=\ln\left(r+\sqrt{r^{2}-1}\right)\), the left-hand side is
\(\displaystyle \sum_{r=2}^{N}\ln\left(r+\sqrt{r^{2}-1}\right).\)
Now integrate by parts:
\(\displaystyle \int_{1}^{N}\cosh^{-1}x\,dx=\left[x\cosh^{-1}x\right]_{1}^{N}-\int_{1}^{N}\frac{x}{\sqrt{x^{2}-1}}\,dx.\)
From part (a),
\(\displaystyle \int_{1}^{N}\frac{x}{\sqrt{x^{2}-1}}\,dx=\left[\sqrt{x^{2}-1}\right]_{1}^{N}=\sqrt{N^{2}-1}.\)
Also \(\cosh^{-1}1=0\), so
\(\displaystyle \int_{1}^{N}\cosh^{-1}x\,dx=N\cosh^{-1}N-\sqrt{N^{2}-1}.\)
Since \(\cosh^{-1}N=\ln\left(N+\sqrt{N^{2}-1}\right)\),
\(\displaystyle \int_{1}^{N}\cosh^{-1}x\,dx=N\ln\left(N+\sqrt{N^{2}-1}\right)-\sqrt{N^{2}-1}.\)
Therefore
\(\displaystyle \sum_{r=2}^{N}\ln\left(r+\sqrt{r^{2}-1}\right)\gt N\ln\left(N+\sqrt{N^{2}-1}\right)-\sqrt{N^{2}-1}.\)
(c) Using left-endpoint rectangles,
\(\displaystyle \cosh^{-1}1+\cosh^{-1}2+\cdots+\cosh^{-1}(N-1)\lt \int_{1}^{N}\cosh^{-1}x\,dx.\)
Since \(\cosh^{-1}1=0\), this becomes
\(\displaystyle \sum_{r=2}^{N-1}\cosh^{-1}r\lt \int_{1}^{N}\cosh^{-1}x\,dx.\)
Adding \(\cosh^{-1}N\) to both sides gives
\(\displaystyle \sum_{r=2}^{N}\cosh^{-1}r\lt \int_{1}^{N}\cosh^{-1}x\,dx+\cosh^{-1}N.\)
Using the value of the integral from part (b),
\(\displaystyle \sum_{r=2}^{N}\cosh^{-1}r\lt (N+1)\cosh^{-1}N-\sqrt{N^{2}-1}.\)
Hence
\(\displaystyle \sum_{r=2}^{N}\ln\left(r+\sqrt{r^{2}-1}\right)\lt (N+1)\ln\left(N+\sqrt{N^{2}-1}\right)-\sqrt{N^{2}-1}.\)
Alternatively, comparing with \(\displaystyle \int_{1}^{N+1}\cosh^{-1}x\,dx\) gives
\(\displaystyle \sum_{r=2}^{N}\ln\left(r+\sqrt{r^{2}-1}\right)\lt (N+1)\ln\left(N+1+\sqrt{N^{2}+2N}\right)-\sqrt{N^{2}+2N}.\)
