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 $\infty$.
 A: put $x = \tan \theta$ given integral is
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sec^2 \theta}{\tan \theta \sec^2 \theta} = \frac{d(\sin \theta)}{\sin \theta} = \ln [|\sin \theta|]^{\frac{\pi}{2}}_{\frac{\pi}{4}} = 0-\ln \frac{1}{\sqrt{2}}=\ln \sqrt{2}$$
A: Why complicate it with complex coefficients?
$$\int_1^\infty\frac{1}{x(x^2+1)}dx=\int_1^\infty\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx=\left(\ln(x)-\frac{1}{2}\ln(x^2+1)\right)\bigg\vert_1^\infty\\=\ln\frac{x}{\sqrt{x^2+1}}\bigg\vert_1^\infty=\ln\frac{1}{\sqrt{1+1/x^2}}\bigg\vert_1^\infty=\ln 1-\ln\frac{1}{\sqrt{2}}=\ln\sqrt 2$$
A: 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
$$ \lim_{x\to \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}} = 1 $$
A: First of all:
$$\frac{1}{x(x^2+1)} = \frac{1}{x}-\frac{1}{2}\left(\frac{1}{x+i} -\frac{1}{x-i}\right)$$
Second of all:
You have to evaluate the integral $\int_1 ^ R\frac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $\infty -\infty$  is undetermined.
$$\int_1 ^ R\frac{1}{x(x^2+1)} = \int_1 ^ R \left[\frac{1}{x}-\frac{1}{2}\left(\frac{1}{x+i} -\frac{1}{x-i}\right)\right]dx$$
$$=[\log |x| - \frac{1}2{}\log(|x+i|) - \frac{1}{2}\log(|x-i|)]_1^R$$
$$=[\log |R| - \frac{1}2{}\log(|R+i|) - \frac{1}{2}\log(|R-i|)] + \frac{1}2{}\log(|1+i|) + \frac{1}{2}\log(|1-i|)]$$
$$=\log\left|\frac{R}{(R+i)^{\frac{1}{2}}(R-i)^\frac{1}{2}}\right|+\frac{1}{2}\log(1+i)(1-i)$$
$$=\log\left|\frac{R}{(R^2+1)^\frac{1}{2}}\right|+\frac{1}{2}\log(2)$$
Now taking the limit yields $\frac{1}{2}\log(2)$.
A: By enforcing the substitution $x=\frac{1}{z}$ the result is straightforward to find:
$$ \int_{1}^{+\infty}\frac{dx}{x(x^2+1)}=\int_{0}^{1}\frac{z\,dz}{z^2+1}=\left[\frac{1}{2}\log(1+z^2)\right]_{0}^{1}=\color{red}{\frac{\log 2}{2}}.$$
