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(a) When $0 \leq x \leq1$, then $x^2 \leq 1$ and $x+1 \geq 1$, so $|f''(x)| = \dfrac{x^2}{|x+1|}\leq \dfrac{1}{1}=1$. (b) To find the maximum value of a function over a closed interval, we test the function's values at the endpoints of the interval and at its critical points inside the interval. The critical points are where the function's derivative is zero or does not exist.

The function we're trying to maximize is $|f''(x)| = \frac{x^2}{|x+1|} = \frac{x^2}{x+1}=f''(x)$ (since our interval only contains nonnegative numbers). So, the critical points occur when $f'''(x) = 0$ or does not exist. We find $f'''(x)$ Using the quotient rule.

$$\begin{aligned} f'''(x)&=\frac{(x+1)(2x)-x^2}{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}\\ 0&=\frac{x(x+2)}{x+1}\\ 0&=x \quad\text{or}\quad x=-1\quad\text{or}\quad x=-2 \end{aligned}$$

The only critical point in $[0,1]$ is $x=0$. So, the extrema of $f''(x)$ over $[0,1]$ will occur at its endpoints. Indeed, since $f'''(x) \geq 0$ for all $x$ in $[0,1]$, $f''(x)$ is increasing over this interval, so its maximum occurs at $x=1$. That is, $$|f''(x)|\leq f''(1)=\frac{1}{2}$$ (c) The absolute error using the midpoint rule is at most $\dfrac{M(b-a)^3}{24n^2}$. Using $M=1$, if we want this to be no more than $10^{-5}$, we find an acceptable value of $n$ with the following calculation:

$$\begin{aligned} \dfrac{M(b-a)^3}{24n^2}&\leq 10^{-5}\\ \dfrac{1}{24n^2}&\leq 10^{-5}&(b-a=1,\,M=1)\\ \frac{10^5}{24} & \leq n^2\\ n & \geq 65 \end{aligned}$$ (d) The absolute error using the midpoint rule is at most $\dfrac{M(b-a)^3}{24n^2}$. Using $M=\frac{1}{2}$, if we want this to be no more than $10^{-5}$, we find an acceptable value of $n$ with the following calculation:

$$\begin{aligned} \dfrac{M(b-a)^3}{24n^2}&\leq 10^{-5}\\ \dfrac{1}{48n^2}&\leq 10^{-5}&(b-a=1,\,M=\frac{1}{2})\\ \frac{10^5}{48} & \leq n^2\\ n & \geq 46 \end{aligned}$$

Remark: how accurate you want to be in these calculations depends a lot on your circumstances. Imagine, for instance, that you were finding $M$ by hand, using this to find $n$ by hand, then programming a computer to evaluate the approximation. For a simple integral like this, the difference between computing time for 65 intervals versus 46 is likely to be miniscule. So, there's not much to be gained by the extra work in (b). However, if your original sloppy $M$ gave you something like $n=1000000$, you might want to put some time into improving it, to shorten computation time. Moreover, if you were finding the approximation by hand, the difference between adding 46 terms and adding 65 terms would be considerable, and you would probably want to put in the effort up front to find the most accurate $M$ possible.

Before we can take our Simpson's rule approximation of $\displaystyle\int_1^x \dfrac{1}{t}~\mathrm{d}t$, we need to know how many intervals to use. That means we need to bound our error, which means we need to bound $\frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}$.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left\{\frac{1}{t}\right\}&=-\frac{1}{t^2}& \frac{\mathrm{d}^2}{\mathrm{d}{t}^2}\left\{\frac{1}{t}\right\}&=\frac{2}{t^3}\\ \frac{\mathrm{d}^3}{\mathrm{d}{t}^3}\left\{\frac{1}{t}\right\}&=-\frac{6}{t^4}& \frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}&=\frac{24}{t^5} \end{aligned}$$

So, over the interval $[1,3]$, $\displaystyle\left|\frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}\right| \leq 24$.

Now, we can find an appropriate $n$ to ensure our error will be be less than 0.1 for any $x$ in $[1,3]$:

$$\begin{aligned} \frac{L(b-a)^5}{180 n^4}&<0 .1 \\\frac{24(x-1)^5}{180 n^4}&< \frac{1}{10} \\n^4 &> \frac{24\cdot (x-1)^5}{18} \\ \text{Because $x-1 \leq 2$ for every $x$ in $[1,3]$, if $n^4 > \dfrac{24\cdot 2^5}{18}$, then $n^4 > \dfrac{24\cdot (x-1)^5}{18}$ for every allowed $x$.} \\ n^4&> \frac{24\cdot 2^5}{18}= \frac{128}{3}\\ n&>\sqrt[4]{\frac{128}{3}}\approx 2.6 \end{aligned}$$

Since $n$ must be even, $n=4$ is enough intervals to guarantee our error is not too high for any $x$ in $[1,3]$. Now we find our Simpson's rule approximation with $n=4$, $a=1$, $b=x$, and $\Delta x = \dfrac{x-1}{4}$. The points where we evaluate $\frac{1}{t}$ are:

$$\begin{aligned} x_0 =& 1 & x_1&=1+ \frac{x-1}{4}& x_2&=1+2\frac{x-1}{4}& x_3&=1+3\frac{x-1}{4}& x_4&=1+4\frac{x-1}{4}\\ & & &=\frac{x+3}{4}& &=\frac{x+1}{2}&&=\frac{3x+1}{4}& &=x \end{aligned}$$

$$\begin{aligned} \log x=\int_1^x \frac{1}{t}~\mathrm{d}t&\approx \frac{\Delta x}{3}\left[\frac{1}{x_0}+\frac{4}{x_1}+\frac{2}{x_2}+\frac{4}{x_3}+\frac{1}{x_4}\right]\\ &=\frac{x-1}{12}\left[1+\frac{16}{x+3}+\frac{4}{x+1}+\frac{16}{3x+1}+\frac{1}{x}\right]\\ &=f(x) \end{aligned}$$

Below is a graph of our approximation $f(x)$ and natural logarithm on the same axes. The natural logarithm function is shown red and dashed, while our approximating function is solid blue. Our approximation appears to be quite accurate for small, positive values of $x$.