5.3 The Fundamental Theorem of Calculus/13

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

${\frac {d}{dx}}\left[h(x)\right]={\frac {d}{dx}}\left[\int _{2}^{1/x}\arctan(t)dt\right]={\frac {-1}{x^{2}}}\cdot (\arctan \left({\frac {1}{x}}\right))-0\cdot (\arctan \left(2)\right)$

=\frac{-\arctan{\frac{1}{x}}</math>

${\text{Therefore, }}g'(x)=$