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 {\displaystyle {\frac {d}{dx}}g(x)={\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=3\cdot {\frac {(3x)^{2}-1}{(3x)^{2}+1}}-2\cdot {\frac {(2x)^{2}-1}{(2x)^{2}+1}}}
= 3 ⋅ 9 x 2 − 1 9 x 2 + 1 − 2 ⋅ 4 x 2 − 1 4 x 2 + 1 {\displaystyle =3\cdot {\frac {9x^{2}-1}{9x^{2}+1}}-2\cdot {\frac {4x^{2}-1}{4x^{2}+1}}} =
= 3 ( 9 x 2 − 1 ) 9 x 2 + 1 − 2 ( 4 x 2 − 1 ) 4 x 2 + 1 {\displaystyle ={\frac {3(9x^{2}-1)}{9x^{2}+1}}-{\frac {2(4x^{2}-1)}{4x^{2}+1}}}
![{\displaystyle {\frac {d}{dx}}g(x)={\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=3\cdot {\frac {(3x)^{2}-1}{(3x)^{2}+1}}-2\cdot {\frac {(2x)^{2}-1}{(2x)^{2}+1}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0fc3ab48595971ace194987825e0d2ea7347dbb5)
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