<s>[INST] Find positive $K$ such that $\int_0^\infty\left(\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}\right)dx$ converges 
Find positive $K$ such that 
  $$\int_0^\infty\left(\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}\right)dx$$ converges

I used the fact that $\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}>\frac{-K+1/\sqrt2}{1+x}$ for $x>1$ to prove it diverges for $K < 1/\sqrt2$.
And the fact that $\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}<\frac{1-K}{x+1}$ to prove it diverges for $K>1$.
However I am not sure how to prove it converges (or not) in $[\frac1{\sqrt2},1]$.
I think it would not converge for any $K$ becase the denomiantor is sorta linear and this would never be identically $0$ for any $K$.
 [/INST] Hint
I suggest to have a look at what happens at the bounds.
For $x$ close to $0$, the Taylor expansion of the integrand is
$$\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}=(1-K)+K x-(K+1) x^2+O\left(x^3\right)$$ 
Now, for infinitely large values of $x$,
$$\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}=\frac{\frac{1}{\sqrt{2}}-K}{x}+\frac{K}{x^2}+O\left(\frac{1}{x^3}\right)$$
Does this tell you something ?
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