<s>[INST] Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$  
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.  
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So
  $$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty  
$$
which evaluates to be $\infty$.
 [/INST] Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's
$$ \frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{1}{2(x+i)} - \frac{1}{2(x-i)} $$
Then 
$$ \int_1^{\infty} f(x) \ dx = \left.\left(\ln |x| - \frac{1}{2}\ln |x+i| -
 \frac{1}{2} \ln |x-i|\right)\right|_1^{\infty} = \left. \ln \left|\frac{x}{\sqrt{x^2+1}} \right|\right|_1^{\infty} = \frac{1}{2}\ln 2 $$
Note that
$$ \lim_{x\to \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}} = 1 $$
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