how does $\displaystyle \int\frac{3x}{3x-2}$ become
$\displaystyle \int \frac{2}{3x-2}+1$
Hello vinson24The short answer is that
$\displaystyle \frac{2}{3x-2}+1 = \frac{2}{3x-2}+ \frac{3x-2}{3x-2}$
$\displaystyle = \frac{2+(3x-2)}{3x-2}$
$\displaystyle =\frac{3x}{3x-2}$
The slightly more difficult trick is to work it the other way around. You do this by algebraic long division, starting with
$\displaystyle 3x-2\quad)\quad 3x = 1$, remainder something.
It's not very easy to set out the working using LaTeX, but it looks something like this:
$\displaystyle \begin{matrix} & &1& & \\ & &--&-&-\\3x-2&)&3x& & \\ & &3x&-&2\\ & &--&-&-\\ & & & &2\end{matrix}$
When you divide $\displaystyle 3x$ by $\displaystyle 3x-2$, then, the quotient is $\displaystyle 1$ and the remainder is $\displaystyle 2$. So the answer can be written:
$\displaystyle \frac{3x}{3x-2} = 1 +\frac{2}{3x-2}$
Grandad