Hallo,
$$f(x) =\frac{1}{4} x^{2}-\ln (\sqrt{x}) \quad 4 \le x \le 9$$
Die Rechnung: $$\begin{aligned} f'\left(x\right)&=\frac{1}{2}x-\frac{1}{2x} \\ L & =\int \limits_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}}\,\text{d}x \\ &= \int \limits_{4}^{9} \sqrt{1+\frac{1}{4}x^2 - \frac{1}{2} + \frac{1}{4x^2}}\,\text{d}x \\ &= \frac{1}{2}\int \limits_{4}^{9} \sqrt{2+x^2 + \frac{1}{x^2}}\,\text{d}x \\ &= \frac{1}{2}\int \limits_{4}^{9} \frac{1}{x}\sqrt{x^4 +2x^2+ 1}\,\text{d}x &&|\, x \ge 0\\ &= \frac{1}{2}\int \limits_{4}^{9} \frac{x^2+1}{x}\,\text{d}x \\ &= \frac{1}{2}\int \limits_{4}^{9} x + \frac{1}{x}\,\text{d}x \\ &= \frac{1}{2}\left[\frac{1}{2}x^2 + \ln(x)\right]_{4}^{9} \\ &= \frac{81-16}{4} +\frac{1}{2}\ln\left(\frac{9}{4}\right) \approx 16,66 \end{aligned} $$und die Bestätigung von Desmos für den nummerischen Wert:
https://www.desmos.com/calculator/sxzjtnxdnp
Gruß Werner