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.

This answer is more for fun than to be taken seriously. I guess you used the conjugate rule trick to calculate the primitive function. Maybe you can use it again to have a simpler expression for the primitive function at the limits?
By the conjugate rule,
$$
\begin{aligned}
\frac{1}{(\sqrt{1+x^2}+x)^2}
&=
\frac{1}{(\sqrt{1+x^2}+x)^2}\frac{(\sqrt{1+x^2}-x)^2}{(\sqrt{1+x^2}-x)^2}\\
&=
(\sqrt{1+x^2}-x)^2\\
&=
(1+x^2)-2x\sqrt{1+x^2}+x^2,
\end{aligned}
$$
so integrating is easy, and a primitive function is given by
$$
F(x)=x+\frac{2}{3}x^3-\frac{2}{3}(1+x^2)^{3/2}.
$$
From the power $3/2$ of $(1+x^2)$, we look for a term
$$
\begin{aligned}
C(\sqrt{1+x^2}-x)^3&=C(1+x^2)^{3/2}-3Cx(1+x^2)+3Cx^2\sqrt{1+x^2}-Cx^3\\
&=4C(1+x^2)^{3/2}-3Cx(1+x^2)-3C\sqrt{1+x^2}-Cx^3.
\end{aligned}
$$
With $C=-1/6$, we can write our $F$ as
$$
\begin{aligned}
F(x)&=x+\frac{2}{3}x^3-\frac{1}{6}(\sqrt{1+x^2}-x)^3-\frac{1}{2}x(1+x^2)-\frac{1}{2}\sqrt{1+x^2}-\frac{1}{6}x^3\\
&=-\frac{1}{6}(\sqrt{1+x^2}-x)^3-\frac{1}{2}(\sqrt{1+x^2}+x)\\
&=-\frac{1}{6}\frac{1}{(\sqrt{1+x^2}+x)^3}-\frac{1}{2}\frac{1}{\sqrt{1+x^2}+x}.
\end{aligned}
$$
From this expression it is easy to see that
$$
\lim_{x\to+\infty}F(x)=0.
$$
