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.

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

$$\begin{align*} \lim_{x\to \infty}\;\frac{2}{3}x^3-\frac{2}{3}\sqrt{(1+x^2)^3}+x&=\lim_{x\to \infty}\; x^3\left(\frac{\dfrac{2}{3}x^3-\dfrac{2}{3}\sqrt{(1+x^2)^3}+x}{x^3}\right) \\ &=\lim_{x\to \infty} x^3\cdot \lim_{x\to \infty}\left(\dfrac{2}{3}-\dfrac{2} {3}\sqrt{\dfrac{(1+x^2)^3}{x^6}}+\frac{1}{x^2}\right)\\ &=\lim_{x\to \infty} x^3\cdot \lim_{x\to \infty}\left(\dfrac{2}{3}-\dfrac{2} {3}\sqrt{\left(\dfrac{1+x^2}{x^2}\right)^3}+\frac{1}{x^2}\right)\\ &=\lim_{x\to \infty} x^3\cdot \left(\frac{2}{3}-\frac{2}{3}\right) \\ &=0 \end{align*}$$

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 $$ \lim_{x\to+\infty}F(x)=0. $$