Find values for the real constants and so that the following limit equation holds:
Incomplete.
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2 comments
Mohammad Azad says:
add and subtract x^2 and x, you get an x and it’s integral is zero, now group the numerator to get (a-1)x^2+(b-1)x and add and subtract
(a-1)x+(a-1) you get (a-1) which is why a should be 1 and then you can add and subtract (b-a)/2, one of the parts will be a logarithm and the other arctan, for the limit to exist a must be 1 and note that the arctanx approaches +-pi/2 as x tends to +- infinity
Anonymous says:
If we do a=1+A and b=1+B yields: x + (Ax^2 + Bx)/(x^2 + x + 1). But x is odd and so we consider only (Ax^2 + Bx)/(x^2 + x + 1).
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and 
so that the following limit equation holds:![\[ \lim_{p \to +\infty} \int_{-p}^p \frac{x^3 + ax^2 + bx}{x^2 + x + 1} \, dx = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6552f0aa7caa6423a568cc25b651fa7a_l3.png "Rendered by QuickLaTeX.com")
add and subtract x^2 and x, you get an x and it’s integral is zero, now group the numerator to get (a-1)x^2+(b-1)x and add and subtract
(a-1)x+(a-1) you get (a-1) which is why a should be 1 and then you can add and subtract (b-a)/2, one of the parts will be a logarithm and the other arctan, for the limit to exist a must be 1 and note that the arctanx approaches +-pi/2 as x tends to +- infinity
If we do a=1+A and b=1+B yields: x + (Ax^2 + Bx)/(x^2 + x + 1). But x is odd and so we consider only (Ax^2 + Bx)/(x^2 + x + 1).