Your substitution is the way to go in this problem. To complete the substitution however, I recommend that you calculate the new limits for $t$. An alternate approach is to postpone this till the end, but this approach invariably ends up confusing if there's a string of substitutions. In this example, as $x \to -\infty$, $t = \arctan x \to -\frac{\pi}{2}$; similarly as $x \to \infty$, we have $t \to \frac{\pi}{2}$. Therefore, after substitution, the integral becomes $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\tan^2 t}} \, dt. $
To proceed further, you need to use the following cousin of the Pythagorean theorem: $ 1 + \tan^2 t = \sec^2 t . $ This identity is extremely important and useful in practice -- not the least in manipulating integrals like this. One should be reminded of this identity whenever one comes across an expression like $\sqrt{1 + \tan^2 t}$ or $\sqrt{1+ x^2}$. By the way, the proof of this identity is based on the standard Pythagorean theorem: $ 1+ \tan^2 t = 1 + \frac{\sin^2 t}{\cos^2 t} = \frac{\cos^2 t + \sin^2 t}{\cos^2 t} = \frac{1}{\cos^2 t} = \sec^2 t. $ From this, it follows that $ \frac{1}{\sqrt{1 + \tan^2 t}} = |\cos t|. $
Thus the integral becomes $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\cos t| \, dt. $ Can you take it from here?