Question

Find: $\int \frac{x^2+1}{x^2-5x+6}$

Answer 1

$\int \frac{x^2+1}{x^2-5x+6}.dx$
$=\int \frac{x^2-5x+6+5x-5}{x^2-5x+6}.dx$
$=\int (\frac{x^2-5x+6}{x^2-5x+6}+\frac{5x-5}{x^2-5x+6}).dx$
$=\int (1+\frac{5x-5}{x^2-5x+6}).dx$
$=\int (1+\frac{5x-5}{(x-2)(x-3)}).dx$
$=\int 1.dx+1\int \frac{5x-5}{(x-2)(x-3)}.dx$
$\stackrel{=x+C_1+\underset{}{\int \frac{5x-5}{(x-2)(x-3)}.dx}}{I}$
Now, finding I
Using partial fraction, we get
$\frac{5(x-1)}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)}$
$\frac{5(x-1)}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}$
$\Rightarrow 5(x-1)=A(x-3)+B(x-2)$
$\Rightarrow 5x-5=(A+B)x-(3A+2B)$
There are two ways to solve this:
Method 1:
Comparing the coefficients of 𝑥 and the constant term, we get
$A+B=5$
$3A+2B=5$
Sovling both, we get
$A=-5,B=10$
Method 2:
using
$5(x-1)=A(x-3)+B(x-2)$
Putting $X=2$
$5(2-1)=A(2-3)+B(2-2$
$\Rightarrow A=-5$
Putting $x=3$
$5(3-1)=A(3-3)+B(3-2)$
$\Rightarrow B=10$
(Usually, second method is easy and faster way to find the variables) Now putting the values of A and B in (2)
$\frac{5(x-1)}{(x-2)(x-3)}=\frac{-5}{(x-2)}+\frac{10}{(x-3)}$
Now,
$I=\int \frac{5(x-1)}{(x-2)(x-3)}.dx$
$=\int \frac{-5}{(x-2)}.dx+\int \frac{10}{(x-3)}.dx$
$\Rightarrow I=-5\log |x-2|+10\log |x-3|+C_2$
$\int \frac{x^2+1}{x^2-5x+6}.dx$
$=x+C_1+I$
$=x+C_1-5\log |x-2|+10\log |x-3|+C_2$
$=x-5\log |x-2|+10\log |x-3|+C$
Where $C=C_1+C_2$