5.3 The Fundamental Theorem of Calculus/53

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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} \\[2ex] &=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{align} }