Substitute [tex]t=\tan s[/tex], so that [tex]\mathrm dt=\sec^2s\,\mathrm ds[/tex]. The integral is then equivalent to
[tex]\displaystyle\int\sqrt{1+(\tan s)^2}\,\sec^2s\,\mathrm ds[/tex]
[tex]=\displaystyle\int\sqrt{1+\tan^2s}\,\sec^2s\,\mathrm ds[/tex]
[tex]=\displaystyle\int\sqrt{\sec^2s}\,\sec^2s\,\mathrm ds[/tex]
In general, [tex]\sqrt{x^2}=|x|[/tex], so [tex]\sqrt{\sec^2s}=|\sec s|[/tex].
We want the substitution made above to be reversible, so that [tex]s=\tan^{-1}t[/tex]. This restricts [tex]s[/tex] to the interval [tex]-\dfrac\pi2<s<\dfrac\pi2[/tex], and over this interval we have [tex]\cos s>0\implies\sec s>0[/tex], so we take the positive square root in order that [tex]\sqrt{\sec^2s}=+\sec s[/tex].
Then the integral becomes
[tex]=\displaystyle\int\sec^3s\,\mathrm ds[/tex]
which can be computed in several ways. One method is to integrate by parts, taking
[tex]u=\sec s\implies\mathrm du=\sec s\tan s\,\mathrm ds[/tex]
[tex]\mathrm dv=\sec^2s\,\mathrm ds\implies v=\tan s[/tex]
so that
[tex]\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s\tan^2s\,\mathrm ds[/tex]
[tex]\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s(\sec^2s-1)\,\mathrm ds[/tex]
[tex]\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int(\sec^3s-\sec s)\,\mathrm ds[/tex]
[tex]\displaystyle2\int\sec^3s\,\mathrm ds=\sec s\tan s+\int\sec s\,\mathrm ds[/tex]
[tex]\displaystyle\int\sec^3s\,\mathrm ds=\frac12\sec s\tan s+\frac12\ln|\sec s+\tan s|+C[/tex]
Then with [tex]s=\tan^{-1}t[/tex], you have [tex]\tan s=t[/tex] and [tex]\sec s=\sqrt{1+t^2}[/tex], which follows from the Pythagorean identity. So
[tex]\displaystyle\int\sqrt{1+t^2}\,\mathrm dt=\frac{t\sqrt{1+t^2}+\ln|t+\sqrt{1+t^2}|}2+C[/tex]
