Answer
$$\displaystyle{\int_0^2\frac{1}{\sqrt{4+t^2}}dt=\ln\left|\sqrt{2}+1\right|}\\$$
Work Step by Step
$\displaystyle{I=\int_0^2\frac{1}{\sqrt{4+t^2}}dt}\\$
$\displaystyle \left[\begin{array}{ll} t=2\tan\theta & t^2=4\tan^2\theta \\ & \\ \frac{dt}{d\theta}=2\sec^2\theta & dt=2\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_0^{\frac{\pi }{4}}\frac{1}{\sqrt{4+4\tan^2}}2\sec^2\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi }{4}}\frac{1}{\sqrt{4\left(1+\tan^2\theta\right)}}2\sec^2\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi }{4}}\frac{1}{2\sec\theta}2\sec^2\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi }{4}}\sec\theta\ d\theta}\\
\displaystyle{I=\left[\ln\left|\sec\theta+\tan\theta\right|\right]_0^{\frac{\pi }{4}}}\\
\displaystyle{I=\ln\left|\sqrt{2}+1\right|-\ln\left|1+0\right|}\\
\displaystyle{I=\ln\left|\sqrt{2}+1\right|}\\
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