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>

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