g ( x ) = ∫ 2 x 3 x u 2 − 1 u 2 + 1 d u {\displaystyle g(x)=\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du}
d d x [ g ( x ) ] = d d x [ ∫ 2 x 3 x u 2 − 1 u 2 + 1 d u ] = 3 ⋅ ( 3 x ) 2 − 1 ( 3 x ) 2 + 1 − 2 ⋅ ( 2 x ) 2 − 1 ( 2 x ) 2 + 1 = 3 ⋅ 9 x 2 − 1 9 x 2 + 1 − 2 ⋅ 4 x 2 − 1 4 x 2 + 1 = 3 ( 9 x 2 − 1 ) 9 x 2 + 1 − 2 ( 4 x 2 − 1 ) 4 x 2 + 1 {\displaystyle {\begin{aligned}{\frac {d}{dx}}[g(x)]&={\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]\\[2ex]&=3\cdot {\frac {(3x)^{2}-1}{(3x)^{2}+1}}-2\cdot {\frac {(2x)^{2}-1}{(2x)^{2}+1}}=3\cdot {\frac {9x^{2}-1}{9x^{2}+1}}-2\cdot {\frac {4x^{2}-1}{4x^{2}+1}}\\[2ex]&={\frac {3(9x^{2}-1)}{9x^{2}+1}}-{\frac {2(4x^{2}-1)}{4x^{2}+1}}\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}{\frac {d}{dx}}[g(x)]&={\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]\\[2ex]&=3\cdot {\frac {(3x)^{2}-1}{(3x)^{2}+1}}-2\cdot {\frac {(2x)^{2}-1}{(2x)^{2}+1}}=3\cdot {\frac {9x^{2}-1}{9x^{2}+1}}-2\cdot {\frac {4x^{2}-1}{4x^{2}+1}}\\[2ex]&={\frac {3(9x^{2}-1)}{9x^{2}+1}}-{\frac {2(4x^{2}-1)}{4x^{2}+1}}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/03f5e4eae250d1091b32a85ab9644557d431daea)