Calculate integral from 1 to ² (ln x)/x dx with accuracy

Calculate $\displaystyle\int_1^2\frac{\ln x}{x}\,dx$ with accuracy $0.01$.

Step 1: Exact value

With $u=\ln x,\ du=dx/x$: $\displaystyle\int\frac{\ln x}{x}\,dx=\frac{(\ln x)^2}{2}+C$, so

\[I_{\text{exact}}=\left[\frac{(\ln x)^2}{2}\right]_1^2=\frac{(\ln2)^2}{2}\approx0.240227.\]

Step 2: Bound the fourth derivative

Differentiating $f(x)=(\ln x)/x$ repeatedly gives $f^{(4)}(x)=\dfrac{24\ln x-50}{x^5}$, which is largest in magnitude at $x=1$: $|f^{(4)}(1)|=50$.

Step 3: Determine the number of subintervals

\[\frac{1}{180(2n)^4}\times50\le0.01\ \Longrightarrow\ n^4\ge1.74\ \Longrightarrow\ n\ge2.\]

So $2n=4$ subintervals suffice.

Step 4: Apply Simpson's formula

$h=0.25$, nodes $x=1,1.25,1.5,1.75,2$; $f(1)=0$, $f(1.25)\approx0.178515$, $f(1.5)\approx0.270310$, $f(1.75)\approx0.319780$, $f(2)\approx0.346574$. Weighted sum $S\approx2.880375$:

\[I_{\text{Simpson}}=\frac{0.25}{3}\times2.880375\approx0.240031.\]

\[\boxed{I_{\text{Simpson}}\approx0.240031}\]

Step 5: Accuracy check

\[E=|0.240227-0.240031|\approx1.95\times10^{-4}<0.01,\]

well within the required tolerance (about 50 times better than needed), confirming the exact value $\dfrac{(\ln2)^2}{2}$ from Step 1.