Question

Solve integration x minus 1 divided by (x minus 2) (x minus 3 )

Answer 1

To find the value of,

$\displaystyle\longrightarrow I=\int\dfrac{x-1}{(x-2)(x-3)}\ dx$

We need to perform partial fraction decomposition.

Let,

$\longrightarrow\dfrac{x-1}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}$

$\longrightarrow A(x-3)+B(x-2)=x-1$

Put $x=3,$ then,

$\longrightarrow B=2$

Put $x=2,$ then,

$\longrightarrow -A=1$

$\longrightarrow A=-1$

So,

$\longrightarrow\dfrac{x-1}{(x-2)(x-3)}=-\dfrac{1}{x-2}+\dfrac{2}{x-3}$

Therefore,

$\displaystyle\longrightarrow I=\int\left(-\dfrac{1}{x-2}+\dfrac{2}{x-3}\right)\ dx$

$\displaystyle\longrightarrow I=-\int\dfrac{1}{x-2}\ dx+2\int\dfrac{1}{x-3}\ dx$

$\displaystyle\longrightarrow I=-\log|x-2|+2\log|x-3|+C$

$\displaystyle\longrightarrow\underline{\underline{I=\log\left|\dfrac{(x-3)^2}{x-2}\right|+C}}$