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.
 A: It is easy to evaluate that definite integral by successive substitutions. I don't see why you evaluate the indefinite one first.

\begin{align}\int_0^\infty \frac{\mathrm dx}{(x+\sqrt{1+x^2})^2}&=\int_0^{\pi/2} \frac{\sec^2 u}{(\tan u+\sec u)^2}\mathrm du \text{ ,via $x=\tan u$}\\&=\int_0^{\pi/2}\frac{\mathrm du}{(\sin u+1)^2}\\&=2\int_0^1 \frac{1+t^2}{(1+t)^4}\mathrm dt\text{ ,via $t=\tan \frac{u}{2}$}\\&=2\int_1^2 \frac{(w-1)^2+1}{w^4}\mathrm dw\text{ ,via $w=t+1$}\\&=2\int_1^2 \left(\frac{1}{w^2}-\frac{2}{w^3}+\frac{2}{w^4}\right)\mathrm dw\\&=\frac{2}{3}\end{align}
Footnote
Weierstrass substitution
A: 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
$$
\lim_{x\to+\infty}F(x)=0.
$$
A: It's easy to forget about the less common hyperbolic substitutions, but when you do use them you appreciate them even more. Let $x = \sinh t$
$$I = \int_0^\infty \frac{\cosh t \:dt}{(\sinh t + \cosh t)^2} = \int_0^\infty e^{-2t}\left(\frac{e^t+e^{-t}}{2}\right)\:dt = \frac{1}{2}+\frac{1}{6} = \boxed{\frac{2}{3}}$$
A: Let’s tackle the limit by L’Hospital Rule and Rationalization.
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[2 x^{3}+3 x-2\left(1+x^{2}\right)^{\frac{3}{2}}\right] \\
=& \lim _{x \rightarrow \infty} \frac{2+\frac{3}{x^{2}}-2\left(\frac{1}{x^{2}}+1\right)^{\frac{3}{2}}}{\frac{1}{x^{3}}} \quad \left(\frac{0}{0}\right) \\
=& \lim _{x \rightarrow \infty} \frac{-\frac{6}{x^{3}}-3 \sqrt{\frac{1}{x^{2}}+1}\left(-\frac{2}{x^{3}}\right)}{-\frac{3}{x^{4}}} \\
=& 2 \lim _{x \rightarrow \infty} \frac{\left(x-\sqrt{1+x^{2}}\right)\left(x+\sqrt{1+x^{2}}\right)}{x+\sqrt{1+x^{2}}} \\
=&-2 \lim _{x \rightarrow \infty} \frac{1}{x+\sqrt{1+x^{2}}} \\
=& 0
\end{aligned}
$$
