886. Find the mass of the arc $A B$ of the curve $y=\ln x$, if at each point of it the linear density is proportional to the square of the abscissa of the point; $x_{A}=1, x_{B}=3$.
Solution. We apply formula (3). Based on the given equation of the curve, we transform the curvilinear integral in formula (3) into an ordinary one with the variable $x$:

$$
\begin{gathered}
y^{\prime}=\frac{1}{x}, \quad d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x=\sqrt{1+\frac{1}{x^{2}}} d x, \quad \delta=k x^{3} \\
m=\int_{\dot{A} B} \delta d l=k \int_{1}^{3} x^{2} \sqrt{\frac{x^{2}+\overline{1}}{x^{2}}} d x=\frac{k}{2} \int_{1}^{3}\left(x^{2}+1\right)^{\frac{1}{2}} d\left(x^{2}+1\right)= \\
=\left.\frac{k}{3}\left(x^{2}+1\right)^{\frac{3}{2}}\right|_{1} ^{3}=\frac{k}{3}(10 \sqrt{10}-2 \sqrt{2}) \approx 9.6 k
\end{gathered}
$$