<s>[INST] For what values of $k$ does the integral $\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$ converge? For what values of $k$ does the integral $$\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$$ converge?
I believe it does not converge for any $k$. 
 [/INST] $\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}=\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}+\int_{1}^{2} \frac{dx}{x^k \sqrt{x+x^2}}+\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$
If $0<x<1$,  $\sqrt{x+x^2}\ge \sqrt{x}$ hence $\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}\le \int_{0^+}^{1} x^{-k-\frac{1}{2}} dx<\infty$ if $-k+\frac{1}{2}>0$, that is $k<\frac{1}{2}$.
If $x>2$,  $\sqrt{x+x^2}\ge x$ hence $\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}\le \int_{2}^{\infty} x^{-k-1} dx<\infty$ if $-k<0$, that is $k>0$.
So it converges for $0<k<\frac{1}{2}$.
If $k\ge \frac{1}{2}$,$\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}=\infty$ while if $k\le 0$, $\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}=\infty$.
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