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.
Best Answer
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