Answer:
[tex]\frac{arctan^2(x)}{2}[/tex]
Step-by-step explanation:
For this question, set u = arctan(x). This would be the easiest way because the derivative of arctan(x) is [tex]\frac{1}{x^2+1}[/tex] which is what we have. Before setting u, always look at the question and think about possible derivatives.
[tex]\int\frac{arctanx}{x^2+1}dx \ Let \ u = arctan(x)\\\\\\u = arctan(x) \ \ \ du = \frac{1}{x^2+1}dx\\[/tex]
Next, plug in u and du.
[tex]\int (arctan(x)*\frac{1}{x^2+1})dx[/tex] (rearranging)
[tex]\int u\ du[/tex]
[tex]= \frac{u^2}{2} = \frac{1}{2}u^2[/tex]
Substitute arctan(x) back in for u and add +C.
[tex]\frac{arctan^2(x)}{2}+C[/tex]
