5.3 The Fundamental Theorem of Calculus/13

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

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}\left[h(x)\right]={\frac {d}{dx}}\left[\int _{2}^{1/x}\arctan(t)dt\right]={\frac {-1}{x^{2}}}\cdot \left(\arctan \left({\frac {1}{x}}\right)\right)-0\cdot (\arctan \left(2)\right)={\frac {-\arctan {\frac {1}{x}}}{x^{2}}}}

${\text{Therefore, }}h'(x)={\frac {-\arctan {\frac {1}{x}}}{x^{2}}}$