1296. Calculate the length of the arc of the parabola $y^{2}=4 x$ from the vertex to the point $M(1 ; 2)$.
Solution. Apply formula (2). From the parabola equation $x=\frac{y^{2}}{4}, x^{\prime}=\frac{y}{2}$, by the condition $y$ varies from 0 to 2, therefore,

$$
s=\int_{0}^{2} \sqrt{1+\frac{y^{2}}{4}} d y=\frac{1}{2} \int_{0}^{2} \sqrt{4+y^{2}} d y
$$

It is known that $\int \sqrt{x^{2}+a} d x=\frac{x}{2} \sqrt{x^{2}+a}+\frac{a}{2} \ln \left|x+\sqrt{x^{2}+a}\right|$ (see the answer to problem 1140). We will use this result:

$$
\begin{aligned}
& s=\left.\frac{1}{2}\left[\frac{y}{2} \sqrt{y^{2}+4}+\frac{4}{2} \ln \left(y+\sqrt{y^{2}+4}\right)\right]\right|_{0} ^{2}= \\
& =\sqrt{2}+\ln (2+2 \sqrt{2})-\ln 2=\sqrt{2}+\ln (1+\sqrt{2})
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-251.jpg?height=406&width=515&top_left_y=443&top_left_x=58)

Fig. 83

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-251.jpg?height=424&width=536&top_left_y=423&top_left_x=583)

Fig. 84