5.3 The Fundamental Theorem of Calculus/53: Difference between revisions

Revision as of 16:00, 26 August 2022

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

${\frac {d}{dx}}\left[\int _{2x}^{3x}{\frac {u^{2}-1}{u^{2}+1}}du\right]=(3*{\frac {3x^{2}-1}{3x^{2}+1}})-(2*{\frac {2x^{2}-1}{2x^{2}+1}}$

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	<math>\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du</math>		<math>\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du</math>

	<math>\frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3*\frac{3x^2-1}{3x^2+1})</math>		<math>\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}</math>