MIT Integration Bee 2023

Evaluate the indefinite integral

\[ \int\bigg(\sqrt{x+1}-\sqrt{x}\bigg)^{\pi}dx \]

First, we can substitute the thing inside the bracket with \(u\). Let \(u= \sqrt{x+1}-\sqrt{x}\), then

\[ \frac{du}{dx} =\frac12\bigg(\frac1{\sqrt{x+1}}-\frac1{\sqrt x}\bigg)=-\frac{u}{2\sqrt{x(x+1)}} \]

Also, we notice that

\[ u^2 = 1 - 2\sqrt{x(x+1)},u^{-2}= 1+2\sqrt{x(x+1)} \]

This implies

\[ \sqrt{x(x+1)}=\frac{u^{-2}-u^2}{4}\implies \frac{du}{dx}=\frac{2u}{u^2-u^{-2}}\\ dx=\frac{1}{2}(u+u^{-3})du \]

Thus,

\[ \begin{aligned} \int\bigg(\sqrt{x+1}-\sqrt{x}\bigg)^{\pi}dx &=\frac12\int(u^{\pi+1}-u^{\pi-3})du\\ &=\frac12\bigg(\frac{u^{\pi+2}}{\pi+2}-\frac{u^{\pi-2}}{\pi - 2}\bigg)+C\\ &=\frac12\bigg(\frac{(\sqrt{x+1}-\sqrt x)^{\pi+2}}{\pi + 2}-\frac{(\sqrt{x+1}-\sqrt x)^{\pi-2}}{\pi - 2}\bigg)+C\\ \end{aligned} \]