5.3 The Fundamental Theorem of Calculus/53: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 1: | Line 1: | ||
<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})-(2*\frac{2x^2-1}{2x^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> | ||
<math>=(3*\frac{9x^2-1}{9x^2+1})-(2*\frac{4x^2-1}{4x^2+1})</math>= | |||
<math>=(\frac{3(9x^2-1)}{9x^2+1})-(\frac{2(4x^2-1)}{4x^2+1})</math> | |||
Revision as of 16:09, 26 August 2022
![{\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)
=
