How to do partial fraction decomposition.

Question

I have a to solve.$$\int\dfrac{1}{(a-2x)^2(b-x)}dx$$ But I don't even know how to start it. Please help.

Answer 1

Starting with $$ \dfrac{1}{(a-2x)(b-x)}=\dfrac{1}{a-2b}\left(\dfrac{-2}{a-2x}+\dfrac{1}{b-x}\right), $$ we get \begin{eqnarray} \dfrac{1}{(a-2x)^2(b-x)}&=&\dfrac{1}{a-2b}\left[\dfrac{-2}{(a-2x)^2}+\dfrac{1}{(a-2x)(b-x)}\right]\\ &=&\dfrac{1}{a-2b}\left[\dfrac{-2}{(a-2x)^2}+\dfrac{1}{a-2b}\left(\dfrac{-2}{a-2x}+\dfrac{1}{b-x}\right)\right]\\ &=&\dfrac{1}{2b-a}\cdot\dfrac{2}{(2x-a)^2}+\dfrac{1}{(2b-a)^2}\cdot\dfrac{2}{2x-a}-\dfrac{1}{(2b-a)^2}\cdot\dfrac{1}{x-b}. \end{eqnarray} It follows that \begin{eqnarray} \int\dfrac{1}{(a-2x)^2(b-x)}\,dx&=&\int\left[\dfrac{1}{2b-a}\cdot\dfrac{2}{(2x-a)^2}+\dfrac{1}{(2b-a)^2}\cdot\dfrac{2}{2x-a}-\dfrac{1}{(2b-a)^2}\cdot\dfrac{1}{x-b}\right]\,dx\\ &=&\dfrac{-1}{2b-a}\cdot\dfrac{1}{2x-a}+\dfrac{1}{(2b-a)^2}\cdot\ln|2x-a|-\dfrac{1}{(2b-a)^2}\cdot\ln|x-b|+c \end{eqnarray}