5.3 The Fundamental Theorem of Calculus/53

$g(x)=\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du$

${\begin{aligned}{\frac {d}{dx}}[g(x)]&={\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]\\[2ex]&=3\cdot {\frac {(3x)^{2}-1}{(3x)^{2}+1}}-2\cdot {\frac {(2x)^{2}-1}{(2x)^{2}+1}}=3\cdot {\frac {9x^{2}-1}{9x^{2}+1}}-2\cdot {\frac {4x^{2}-1}{4x^{2}+1}}\\[2ex]&={\frac {3(9x^{2}-1)}{9x^{2}+1}}-{\frac {2(4x^{2}-1)}{4x^{2}+1}}\\[2ex]\end{aligned}}$