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I'm having a bit of trouble with calculating the value of the following improper integral via integration by parts...

\begin{equation} \int_1^{+\infty} \frac{\arctan x}{x^2} \,\, dx \end{equation}

I have tried to do the following but I got stuck with calculating the integral of the righthand side...

\begin{equation} \int_a^b \frac{\arctan x}{x^2} \,\, dx = -\left.\frac{\arctan x}{x}\right|_a^b + \int_a^b -\frac{1}{x}\cdot\frac{1}{1+x^2} \,\,dx \end{equation}

Can anyone kindly point me out where to go from here? I, of course, know how to get out of here once I can determine

\begin{equation} \int_a^b -\frac{1}{x}\cdot\frac{1}{1+x^2} \,\,dx \end{equation}

but I don't seem to be able to compute this. Thank you very much.

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@Argon: Yes, and that results in 1/x-x/(x^2+1)... Is that right? From there I think I can reach it with logarithms... Right? –  José Fonseca Jan 26 '13 at 23:43

1 Answer 1

$$\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}$$

Now integrate both individually.

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Precisely what I have recalled. Thank you :-) –  José Fonseca Jan 26 '13 at 23:44
    
@JoséFonseca Sorry, I deleted my comment the second before you replied! But yes, you are correct. –  Argon Jan 26 '13 at 23:45
    
No problem @Argon :-) Thank you very much :-) –  José Fonseca Jan 26 '13 at 23:56

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