Question 1182679
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^\infty\,\frac{dx}{\(2x\,+\,1\)^2}]


Let *[tex \Large u\ =\ 2x\ +\ 1] then *[tex \Large \frac{du}{dx}\ =\ 2] so *[tex \Large dx\ =\ \frac{1}{2}du]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\frac{1}{\(2x\,+\,1\)^2}\,dx\ =\ \frac{1}{2}\int\,\frac{1}{u^2}\,du\ =\ -\frac{1}{2u}]


By the Power Rule.  Now substitute back:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\frac{1}{\(2x\,+\,1\)^2}\,dx\ =\ -\frac{1}{4x\,+\,2}\ +\ C]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^\infty\,\frac{dx}{\(2x\,+\,1\)^2}\ =\ \lim_{t\right\infty}\,\int_0^t\,\frac{dx}{\(2x\,+\,1\)^2}\ =\ \lim_{t\right\infty}\,\(\(-\frac{1}{4t\,+\,2}\)\ -\ \(-\frac{1}{4(0)\,+\,2}\)\)\ =\ \frac{1}{2}]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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