SOLUTION: The integral of the reciprocal of the square of (2x + 1), with respect to x from zero to infinity is?
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Question 1182679: The integral of the reciprocal of the square of (2x + 1), with respect to x from zero to infinity is?
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
^2})
Let 
then 
so 
^2}\,dx\ =\ \frac{1}{2}\int\,\frac{1}{u^2}\,du\ =\ -\frac{1}{2u})
By the Power Rule. Now substitute back:
^2}\,dx\ =\ -\frac{1}{4x\,+\,2}\ +\ C)
Then
^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
My calculator said it, I believe it, that settles it
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