Answer
$$
\begin{aligned}
\int_{1}^{\sqrt{3}} \frac{\sqrt{1+x^{2}}}{x^{2}} d x&=\sqrt{2}-\frac{2}{\sqrt{3}}+\ln (2+\sqrt{3})-\ln (1+\sqrt{2})
\end{aligned}
$$
Work Step by Step
Given
$$
\int_{1}^{\sqrt{3}} \frac{\sqrt{1+x^{2}}}{x^{2}} d x
$$
Let
$$
x=\tan \theta, \text { so that } d x=\sec ^{2} \theta d \theta,\\
$$
$$
x=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}, \text { and } x=1 \Rightarrow \theta=\frac{\pi}{4} .
$$
Then
$$
\begin{aligned}
\int_{1}^{\sqrt{3}} \frac{\sqrt{1+x^{2}}}{x^{2}} d x&=\int_{\pi / 4}^{\pi / 3} \frac{\sec \theta}{\tan ^{2} \theta} \sec ^{2} \theta d \theta \\
&=\int_{\pi / 4}^{\pi / 3} \frac{\sec \theta\left(\tan ^{2} \theta+1\right)}{\tan ^{2} \theta} d \theta \\
&=\int_{\pi / 4}^{\pi / 3}\left(\frac{\sec \theta \tan ^{2} \theta}{\tan ^{2} \theta}+\frac{\sec \theta}{\tan ^{2} \theta}\right) d \theta\\
& =\int_{\pi / 4}^{\pi / 3}(\sec \theta+\csc \theta \cot \theta) d \theta \\
&=[\ln |\sec \theta+\tan \theta|-\csc \theta]_{\pi / 4}^{\pi / 3} \\
&=\left(\ln |2+\sqrt{3}|-\frac{2}{\sqrt{3}}\right)-(\ln |\sqrt{2}+1|-\sqrt{2}) \\
&=\sqrt{2}-\frac{2}{\sqrt{3}}+\ln (2+\sqrt{3})-\ln (1+\sqrt{2}).
\end{aligned}
$$
