I rmb replying to this post. Not sure what's taking so long for my post to appear? ...maybe I didn't reply in the end.
Anyway, you can write (arctanx)/x as a power series. It's a known series. Just use Maclaurin's Series, which is the power series expression of a function f(x) = f(0) + (x)f'(0)/1! + (x^2)f''(0)/2! + ...
One easy way is to start with arctanx = x - (x^3)/3 + (x^5)/5 + ...
Then divide each term by x.
Then notice each term is just a polynomial expression.
Ur answer is just to write the integral of each term summed together.
i.e. integral((arctanx)/x) = integral(1 - (x^2)/3 + (x^4)/5 +...) = integral(1) - integral((x^2)/3) + integral((x^4)/5) + ... = x - (x^3)/9 + ... (you get the idea.)
---------- Post added at 01:30 PM ---------- Previous post was at 01:23 PM ----------
Sorry about how late my reply is. Doubt my answer is very useful for your homework by now. But this topic is widely used so you might use it again if you're studying university calculus.
The radius of convergence just determines what interval the power expansion exists for a function.
Specifically, the Maclaurin Series only exists for functions where |x|<R = radius of convergence. Note that the Maclaurin Series is a special case of the Taylor Series of a function at a. In the Maclaurin Series, a = 0 and c_n = f^(n) (a) / n!
---------- Post added at 01:34 PM ---------- Previous post was at 01:30 PM ----------
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