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f(a) =ln a, to belong to the real has as the only constraint: a>0. In your case: f(x) = ln(1+x^2); so the only limitation is: (1+x^2)>0; You can see that the parenthesis is a parabola quadratic vertex lower (quadratic term positive), and that its roots are not real: 1+x^2=0; x^2=-1, x= +-√ -1; x=+-i; This indicates that never crosses the x-axis (or what is the same: for any value of x will be positive or>0). In short, your function is valid for all x=R; or: (-∞; +∞).
When I have questions about the domain of a function, I see your graph, I enter the website online, symbolab, it is free and do not need to register necessarily. In your case if entrás to that site, you only have to write in the following way: y=ln(1+x^2). The site symbolab is a dot com site. There you enter the expression, as I just did recently and it gives the domain and range(image) and other data, you will also see the graph. For your expression the domain is all real numbers, or as you said albert buscapolos: the domain is from - infinity to + infinity. But in any graphing software, such as geogebra, or another, or one for android, enter the expression and look on the graph and you'll notice that x-axis values have their corresponding point plotted in the plane, and which are not (to exclude those from the domain), in this case all the values of x have a point graphed on the plane, that is to say that you have your image all values of x. Or is that all x are part of the domain. The domain are all real, in your mathematical function. With graph as well, or by hand, always by observing the graph you will notice the domain. Looking if to each point of the x-axis corresponds to a point plotted in the plane, either up or down or about the same point on the x-axis. Look at each point on the x-axis and you fijás if just up or down or about the same corresponds to a point plotted and the value of x belongs to the domain of the function.