Calculating Integral with Series Form of Tan-1x: Step-by-Step Guide

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SUMMARY

The discussion focuses on calculating the integral \(\int \frac{x - \tan^{-1} x}{x^3} \, dx\) using the series form of \(\tan^{-1} x\), which is defined as \(\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}\). Participants suggest separating the integral into simpler components for easier evaluation, particularly emphasizing the need to integrate \(-\tan^{-1}(x)/x^3\) and correctly adjusting the series representation. The correct approach involves recognizing that dividing by \(x^3\) reduces the power of \(x\) by three, leading to a more manageable series form.

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[tex]\int \frac{x-\tan^{-1} x}{x^3}[/tex]

I know the series form of tan-1 x = [tex]\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}[/tex]

I know I need to subtract the x from that series and divide the x cubed form that series but i can't seem to be able to right the resulting series in a general form, any hints?

I thought it would be: [tex]-1/3x + \sum_{n=2}^{\infty} -x^{2n+2}/(2n+1)(2n+2)[/tex]. But this isn't it. Any Help?
 
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I don't see how you arrived at that series. What you may want to consider doing also, for simplicity's sake, is to separate the integral into one that can be easily evaluated and one that should be done by series. Redo your solution paying particular attention to integrating -arctan(x)/x^3
 
Dividing by x^3 just reduces the power of x by 3, ie, x^n/x^3=x^(n-3). Your answer is close.
 

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