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I been trying to get the following $\frac{1}{ax+b}$ on the form $Ax^{-1}+B$ but I cant get it to work. Any ideas?

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    Is there a reason you're trying to do this? I'm not sure it's possible... – 2017-02-25
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    I try to understand rational functions, and just tried it – 2017-02-25
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    Cannot be done unless $b=0$. Let $x$ be tiny, very close to zero. Then $Ax^{-1}$ will be huge (in terms of absolute value). But $1/(ax+b)$ will be close to $1/b$. – 2017-02-25

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Algebraically, you can try like this:

Given $\dfrac {1}{ax+b}$, suppose you could find constants $A, B$ such that for all $x$(except $-\dfrac {b}{a}$), $\dfrac {1}{ax+b} = \dfrac Ax + B$. Then you could multiply both sides by $x(ax+b)$ and get $x = A(ax+b) + Bx(ax+b)$, or $aBx^2+ (Aa+Bb-1)x + Ab = 0$. But this is a quadratic equation, and we know it only has at most $2$ solutions. Therefore we have a contradiction, because we claimed our equation holds for infinitely many $x$.

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This isn't possible for nonzero $b$. $\dfrac {1}{ax+b} = \dfrac 1a \cdot \dfrac {1}{x + \frac ba}$. Thinking about this graphically, this is a vertical stretch of $\dfrac 1a$ and a horizontal shift of $\dfrac ba$ of the graph of $\dfrac 1x.$

You are trying to get this as $A \cdot \dfrac 1x + B$, wich is some vertical stretch of $A$ and a vertical shift of $B$ of the graph of $\dfrac 1x$. There's no horizontal shift, so there's no way to ever match it up with $\dfrac 1a \cdot \dfrac {1}{x + \frac ba}$