# Determine whether this integral converges: $\int_1^\infty\frac{(x+1)\arctan x}{(2x+5)\sqrt x}$

Determine whether the next integral converges: $$\int_1^\infty\frac{(x+1)\arctan x}{(2x+5)\sqrt x}$$

I has this one on a test and lost all my points on this one. Since we were given no answers to the test I still have no idea how to solve it.

Can you please give me the idea on how to solve that one?

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Note that $$\lim_{x\to\infty} \arctan x=\frac{\pi}{2}$$ so $$\frac{(x+1)\arctan x}{(22x+5)\sqrt x}\sim_\infty\frac{\pi}{44\sqrt{x}}$$ and since the integral $$\int_1^\infty \frac{dx}{\sqrt{x}}$$ is divergent hence the given integral is also divergent by limit comparaison.
Just as a constructive criticism, I would have started with $$\lim_{x\to\infty}\frac{(x+1)\arctan(x)}{22x+5}=\frac\pi{44}$$ then find $M$ so that when $x\ge M$, $\frac{(x+1)\arctan(x)}{22x+5}\ge\frac1{44}$. Then you can compare with $$\int_M^\infty\frac{\mathrm{d}x}{44\sqrt{x}}$$ –  robjohn Jun 13 '13 at 20:52
By the integral test, the convergence of given integral is equivalent to the convergence of $$\sum_{n=1}^\infty\frac{(n+1)\arctan n}{(22n+5)\sqrt n}$$ Also notice that $$\lim_{n\rightarrow\infty}\frac{ \frac{(n+1)\arctan n}{(22n+5)\sqrt n} }{\frac{1}{\sqrt n}}=\frac{\pi}{44}$$ Thus by the limit comparison test, since $\sum\frac{1}{\sqrt n}$ diverges, given integral also diverges.