∫ 2 x 3 x u 2 − 1 u 2 + 1 d u {\displaystyle \int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du}
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}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=(3*{\frac {3x^{2}-1}{3x^{2}+1}})-(2*{\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*{\frac {9x^{2}-1}{9x^{2}+1}})-(2*{\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}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=(3*{\frac {3x^{2}-1}{3x^{2}+1}})-(2*{\frac {2x^{2}-1}{2x^{2}+1}})}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/63d1e7666ddb7e3f26b098dbb2c90064bf8335c2)
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