How to work more easily with (ax+b)/(cx+d)

When you’re dealing with an expression like


it is often useful to do to it the algebraic equivalent of rewriting \frac{7}{2} as 3\frac{1}{2} .

Only, what you do here is rewrite the expression in two parts, one of which is just a constant, and the other of which has no “x” in the numerator.

\frac{ax+b}{cx+d} = \frac{a}{c} + \frac{bc-ad}{c(cx+d)}

For example

\frac{x-1}{x+1} = 1 - \frac{2}{x+1}

Now you can differentiate the expression without having to use the quotient rule.

\frac{\mathrm{d}}{\mathrm{d}x} (\frac{x-1}{x+1}) = \frac{2}{(x+1)^2}

You can integrate it without having to worry about whether you should look for a substitution.

\int \frac{x-1}{x+1} \, \mathrm{d}x = x - 2 \ln (x+1) + C

You can sketch it easily. The example is a rectangular hyperbola with one asymptote at y = 1 and another at x = −1, but “upside down” (the bit above the y-asymptote is on the left, not the right).

In general you get a rectangular hyperbola with one asymptote at y = \frac{a}{c} and the other at x = - \frac{d}{c} . Depending on the sign of bc−ad, the hyperbola may or may not be “upside down”.