h ( x ) = ∫ 2 1 / x arctan ( t ) d t {\displaystyle h(x)=\int _{2}^{1/x}\arctan(t)dt}
d d x [ h ( x ) ] = d d x [ ∫ 2 1 / x arctan ( t ) d t ] = − 1 x 2 ⋅ ( arctan ( 1 x ) ) − 0 ⋅ ( arctan ( 2 ) ) = − arctan 1 x x 2 {\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}}}}
Therefore, h ′ ( x ) = − arctan 1 x x 2 {\displaystyle {\text{Therefore, }}h'(x)={\frac {-\arctan {\frac {1}{x}}}{x^{2}}}}
![{\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}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/38499f58d730e54a4a6dbeeda4924be33b09cae5)
