Answer
$$\frac{a}{2} \sqrt{1+4 a^{2}}+\frac{1}{4} \ln |\sqrt{1+4 a^{2}}+2 a|$$
$$1.478943$$
Work Step by Step
Since
\begin{align*}
s&=\int_{a}^{b} \sqrt{1+(f'(x))^{2}} d x\\
&= \int_{0}^{a} \sqrt{1+4 x^{2}} d x
\end{align*}
Let $2x=\tan u ,\ \ 2dx=\sec^2 udu$; then by
$$\int \sec ^{n} u d u=\frac{1}{n-1} \tan u \sec ^{n-2} u+\frac{n-2}{n-1} \int \sec ^{n-2} u d u$$
we get
\begin{align*}
s&=\int_{a}^{b} \sqrt{1+(f'(x))^{2}} d x\\
&= \int_{0}^{a} \sqrt{1+4 x^{2}} d x\\
&= \frac{1}{2}\int_{0}^{\tan^{-1}(2a)}\sec^3 udu\\
&=\left.\left(\frac{1}{4} \sec u \tan u+\frac{1}{4} \ln |\sec u+\tan u|\right)\right|_{ 0} ^{\tan^{-1}(2a)}\\
&=\frac{a}{2} \sqrt{1+4 a^{2}}+\frac{1}{4} \ln |\sqrt{1+4 a^{2}}+2 a|
\end{align*}
Hence, at $a=1$
$$s=\frac{1}{2} \sqrt{5}+\frac{1}{4} \ln (\sqrt{5}+2) \approx 1.478943$$
