<s>[INST] Evaluate $\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}$ How to evaluate this definite integral from MIT Integration Bee 2006?
$$\int_0^\infty \frac{dx}{(x+\sqrt{1+x^2})^2}.$$
So far, I have shown that the indefinite integral is
$$\frac{2x^3 + 3x - 2(1+x^2)^{3/2}}{3}.$$
At $x = 0$, the expression above equals $-\dfrac{2}{3}.$
Using WolframAlpha, I also know that the definite integral equals $\dfrac{2}{3}$.
So the only thing left to show is $$\lim_{x\rightarrow \infty} \frac{2x^3 + 3x - 2(1+x^2)^{3/2}}{3}=0.$$
I'm not sure how to calculate this limit.
 [/INST] Let’s tackle the limit by L’Hospital Rule and Rationalization.
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[2 x^{3}+3 x-2\left(1+x^{2}\right)^{\frac{3}{2}}\right] \\
=& \lim _{x \rightarrow \infty} \frac{2+\frac{3}{x^{2}}-2\left(\frac{1}{x^{2}}+1\right)^{\frac{3}{2}}}{\frac{1}{x^{3}}} \quad \left(\frac{0}{0}\right) \\
=& \lim _{x \rightarrow \infty} \frac{-\frac{6}{x^{3}}-3 \sqrt{\frac{1}{x^{2}}+1}\left(-\frac{2}{x^{3}}\right)}{-\frac{3}{x^{4}}} \\
=& 2 \lim _{x \rightarrow \infty} \frac{\left(x-\sqrt{1+x^{2}}\right)\left(x+\sqrt{1+x^{2}}\right)}{x+\sqrt{1+x^{2}}} \\
=&-2 \lim _{x \rightarrow \infty} \frac{1}{x+\sqrt{1+x^{2}}} \\
=& 0
\end{aligned}
$$
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