Question

The value of the integral $\int \frac{x^2+1}{x^2-5x+6}dx$
(where $C$ is integration constant)

• A. $x-\ln |x+2|+10\ln |x-3|+C$
• B. $x+\ln |x-2|+10\ln |x-3|+C$
• C. $x-5\ln |x-2|+10\ln |x-3|+C$
• D. $x+5\ln |x-2|+10\ln |x-3|+C$

Answer 1

The correct option is C $x-5\ln |x-2|+10\ln |x-3|+C$

Here the integrand $\frac{x^2+1}{x^2-5x+6}$ is not proper rational function, so dividing them, we get
$\frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}=1+\frac{5x-5}{(x-2)(x-3)}$
Let
$\frac{5x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}\Rightarrow 5x-5=A(x-3)+B(x-2)$
Equating the coefficients of $x$ and constant terms on both sides, we get
$A+B=53A+2B=5$
Solving these equations, we get
$A=-5,B=10$
Thus,
$\frac{x^2+1}{x^2-5x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$
Therefore,
$\int \frac{x^2+1}{x^2-5x+6}dx=\int dx-5\int \frac{1}{x-2}dx+10\int \frac{dx}{x-3}=x-5\ln |x-2|+10\ln |x-3|+C$