Question

Solve $\int \frac{x^2+1}{x^2-5x+6}dx$

Answer 1

$\frac{x^2+1}{x^2-5x+6}=\frac{x^2+1-5x+6+5x-6}{x^2-5x+6}$

$=\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}$

$=1+\frac{5x-5}{x^2-5x+6}$

$=1+\frac{5x-5}{x(x-2)-3(x-2)}$

$=1+\frac{5x-5}{(x-2)(x-3)}$

Now solving $\frac{5x-5}{(x-2)(x-3)}$ using partial function

Let $\frac{5(x-1)}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)}$

$\frac{5(x-1)}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)}+\frac{B}{(x-3)}$

$5(x-1)=A(x-3)+B(x-2)$

Putting $x=2$

$5(2-1)=A(2-3)+B(2-2)$

$5=-A+0$

$A=-5$

Similarly putting $x=3$

$5(3-1)=A(3-3)+B(3-2)$

$B=10$

Hence we can write it as

$\frac{5(x-1)}{(x-2)(x-3)}=\frac{-5}{(x-2)}+\frac{10}{(x-3)}$

Now,

$\int \frac{x^2+1}{x^2-5x+6}dx=\int (1+\frac{-5}{(x-2)}+\frac{10}{(x-3)})dx$

$=\int dx-\int \frac{-5}{(x-2)}dx+\int \frac{10}{(x-3)}dx$

$=x-5\log |x-2|+10\log |x-3|+c$