$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1} dx$ diverges? Please, compare and contrast different spaces, it depends on the space! I cannot formulate this mathematically but I think that $\mathbb R$ extended with infinities behave totally differently.
$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1}$ where the term $\int_{0}^{\infty} dx$ diverges.
But
Let's consider $k=1$ then $\int_{0}^{\infty}\frac{1}{x^{2}+1} dx= Arctan(x)_{0}^{\infty} =\frac{\pi}{2}$ so I must have some mistake in 
$$\int_{0}^{\infty}\frac{1}{kx^{2}+1} dx=\int_{0}^{\infty}\frac{(kx^{2}+1)-kx^{2}}{kx^{2}+1} dx=\int_{0}^{\infty} dx -\int_{0}^{\infty}\frac{kx^{2}}{kx^{2}+1}$$
but I cannot see it, where am I doing a mistake?
[Update] 
Let' s take simpler example:
$$\int_{0}^{\infty} \frac{1+x+x^{2}+x^{3}}{x^{k}} dx= \int_{0}^{\infty} x^{-k} dx+\int_{0}^{\infty} x^{-k+1} dx+ \int_{0}^{\infty} x^{-k+2} dx+ \int_{0}^{\infty} x^{-k+3}dx$$
So this is also wrong? I cannot break it up like this? What does it mean to be $indefinite$? If I have extended real space $\mathbb R\cup\{\infty\}\cup\{-\infty\}$, what about now? Now look $\infty$ is just a normal number now, it is not indefinite so does it converge?
 A: $\infty-\infty$ is undefined (the right hand side of your equation).
If $\int_0^\infty f(x)\, dx $ converges and if  $\int_0^\infty g(x)\, dx $ converges, then you can write
$\int_0^\infty[ f(x)+g(x)]\, dx=\int_0^\infty f(x)\, dx +\int_0^\infty g(x)\, dx $.
But you can't split a convergent improper integral into two pieces as you did unless both (or just one) pieces converge, for essentially the reason I gave  at the outset. (Note that both the integrals on the right hand side of your equation are infinite. But you can't do anything with them, the right hand side is not defined, and the step you made going from the left hand side to the right hand side is not justified.)
That $\infty-\infty$ is undefined should become clear if you consider the limits of the expressions: $(n)-(n+1)$, or $(n+17)-(n-42)$, or $(n^2)-(n)$.
A: Suppose $A=B+C$. If the integrals of $A,B,C$ all exist, then $\int A=\int B+\int C$. The trouble is, it is possible that one of the integrals exists and the other two don't. For example, $${1\over3(x-3)}={1\over3x}+{1\over x(x-3)}$$ is an algebraic identity, but if you write $$\int_{-1}^1{dx\over3(x-3)}=\int_{-1}^1{dx\over3x}+\int_{-1}^1{dx\over x(x-3)}$$ you're in trouble because the two integrals on the right don't exist. (Actually, in this case the difficulty already arises before you even integrate; the left side of the identity makes sense for $x=0$, while the right side doesn't.)  
A: You're missing the fact that for $k >0$
$$\int\limits_0^\infty  {\frac{{dx}}{{k{x^2} + 1}} < } \int\limits_0^\infty  {\frac{{dx}}{{{x^2} + 1}} = \frac{\pi }{2}} $$
so the integral converges.
And that putting 
$$\sqrt k x = u$$
$$\int\limits_0^\infty  {\frac{{dx}}{{k{x^2} + 1}}}  = \frac{1}{{\sqrt k }}\int\limits_0^\infty  {\frac{{du}}{{{u^2} + 1}}}  = \frac{1}{{\sqrt k }}\frac{\pi }{2}$$
There is not "mistake" so to call it, but rather that you're not thinking about
$$\int\limits_0^\infty  {f\left( x \right)dx}  = \mathop {\lim }\limits_{u \to \infty } \int\limits_0^u {f\left( x \right)dx} $$
and that when the function is not continuous at $x=c\in(a,b)$ then the integral should be interpreted as:
$$\int\limits_a^b {f\left( x \right)dx}  = \mathop {\lim }\limits_{m \to c} \int\limits_a^m {f\left( x \right)dx}  + \int\limits_m^d {f\left( x \right)dx} $$
In the case of the polynomials you have a discontinuity at $x=0$ so when you take limits you have
$$\int\limits_0^\infty  {{x^{1 - k}}dx}  + \int\limits_0^\infty  {{x^{2 - k}}dx}  + \int\limits_0^\infty  {{x^{3 - k}}dx}  = \left( {\mathop {\lim }\limits_{u \to \infty } \frac{{{u^{2 - k}}}}{{2 - k}} + \frac{{{u^{3 - k}}}}{{3 - k}} + \frac{{{u^{4 - k}}}}{{4 - k}}} \right) - \left( {\mathop {\lim }\limits_{u \to 0} \frac{{{u^{2 - k}}}}{{2 - k}} + \frac{{{u^{3 - k}}}}{{3 - k}} + \frac{{{u^{4 - k}}}}{{4 - k}}} \right)$$
And you can radily check that the limit at infinity is zero for all three expressions but for zero things go to infinity. If $k<0$ then the problem would be at infinity.
