Question

Integrate $\int \frac{x^2+x+2}{(x-1)(x-2)}dx$, with respect to $x$.

Answer 1

$I=\int \frac{x^2+x+2}{(x-1)(x-2)}dx$

$I=\int \frac{x^2-3x+2+4x}{x^2-3x+2}dx=\int 1\frac{4x}{x^2-3x+2}dx$

$I=\int 1.dx+4\int \frac{x}{x^2-3x+2}dx$

$I=I_1+I_2----\left(1\right)$

$I_1=\int 1.dx=x----\left(2\right)$

$I_2=\int \frac{4x}{x^2+3x+2}dx=4\int \frac{x}{(x-1)(x-2)}dx$

using partial fraction we get

$\frac{x}{(x-1)(x-1)}=\frac{A}{x-1}+\frac{B}{x-2}$

$x=A(x-2)+B(x-1)=x(A+B)-2A-B$

on comparing.

A+B = 1

& -2A-B=0 (B = -2A)

$-A=1\Rightarrow$ $[A=-1]$ &$[B=2]$

then

$I_2=4\int \frac{-1}{x-1}+\frac{2}{x-2}dx=4[-log(x-1)+2log(x-2\left)\right]$

$I_2=4\left[log\frac{1}{(x-1)+log(x-2{)}^2}\right]=4log\frac{(x-2{)}^2}{(x-1)}---\left(3\right)$

Then using eq (1),(2) & (3) we have

$I=x+4log\frac{(x-2{)}^2}{(x-1)}+C$