Question

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

Answer 1

Solution:

Before doing partial fraction ,we must divide the given polynomial because here degree of both numerator and denominator are equal.

$x^{2}-5x+6)x^{2}+1(1\\\\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:x^{2}-5x+6\\\\\:\:\:\:\:\:\:\:\:\:\:\:\:5x-5$

So,the expression can be written as

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

Now factorise the denominator

$x^{2}-5x+6=x^{2}-2x-3x+6\\\\x(x-2)-3(x-2)=(x-2)(x-3)$

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

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

second term should be partial factorised for doing integration
$\frac{5(x-1)}{(x-2)(x-3)}= \frac{A}{x-2}+\frac{B}{x-3}\\\\A=\frac{5(x-1)}{(x-3)} put x=2\\\\ A=\frac{-5}{1}\\\\ A= -5\\\\B=\frac{5(x-1)}{(x-2)} put x=3\\\\\\ B=\frac{10}{1}\\\\\\B=10$

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