5.3 The Fundamental Theorem of Calculus/33

$h(x)=\int _{2}^{1/x}\arctan(t)dt$

<math>\frac{d}{dx}\left[h(x)\right]=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]=\frac{-1}{x^2}