Question

if $\int \frac{1}{(a-x)(b-x)}dx=kx$, then find the value of k

Answer 1

Let $\frac{1}{(a-x)(b-x)}=\frac{A}{a-x}+\frac{B}{b-x}$

$\frac{1}{(a-x)(b-x)}=\frac{A(b-x)+B(a-x)}{(a-x)(b-x)}$

$\therefore 1=Ab+Ba-(A+B)x$

On comparing,

$A-B=0$

$-A=-B$____(1)

Also, $Ab+Ba=1$___(2)

Substituting eq.(1) in (2), we get

$\therefore Ab-Aa=1$

$\therefore A=\frac{1}{(a+b)}\frac{1}{(b-a)}$

$\therefore B=-A=\frac{1}{a+b}=\frac{-1}{(b-a)}$

$\therefore \int \frac{1}{(a-x)(b-x)}=\frac{1}{b-a}[\frac{1}{(a-x)}-\frac{1}{(b-x)}]dx$

$=\frac{1}{b-a}[\frac{ln(a-x)}{-1}-\frac{ln(b-x)}{-1}]$

$=\frac{1}{(a+b)}ln\frac{1}{(b-a)}ln\left(\frac{a-x}{b-x}\right)$

on comparing with

$\int \frac{1}{(a-x)(b-x)}=kx,$ we get

$kx=\frac{1}{(b-a)}$ ln $\left(\frac{a-x}{b-x}\right)$

$k=\frac{1}{x(b-a)}$ ln $\left(\frac{a-x}{b-x}\right)$