What is the rational function with a slant asymptote of y = 2x + 1?

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1. Homework Statement

find the rational function with the slant asymptote of y = 2x + 1


2. The attempt at a solution

(2x +1) + something over something
 
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There's no unique rational function. There's tons of them. Start with y=2x+1 and add a fraction that goes to zero as x->infinity (or where ever you want your asymptote).
 
thz i figured it out it sid make up a functions but thanks anyway
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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