COMEDK 2025 Morning Solved Paper

Concept:Using partial fraction decomposition to integrate a rational function with a linear and a quadratic factor.Explanation:Let $\frac{1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}$.Multiplying both sides by $(x+2)(x^2+1)$ gives:$1 = A(x^2+1) + (Bx+C)(x+2)$.Expanding: $1 = (A+B)x^2 + (2B+C)x + (A+2C)$.Equate coefficients:For $x^2$: $A+B=0 \Rightarrow B=-A$For $x$: $2B+C=0 \Rightarrow C = -2B = 2A$For constant: $A+2C=1 \Rightarrow A+4A=5A=1 \Rightarrow A=\frac{1}{5}$.Thus $B=-\frac{1}{5}$, $C=\frac{2}{5}$.Now integrate termwise:$\int \frac{dx}{(x+2)(x^2+1)} = \frac{1}{5}\int \frac{dx}{x+2} - \frac{1}{5}\int \frac{x}{x^2+1}dx + \frac{2}{5}\int \frac{dx}{x^2+1}$.Calculate each integral:$\int \frac{dx}{x+2} = \ln|x+2|$$\int \frac{x}{x^2+1}dx = \frac{1}{2}\ln(x^2+1)$ (since derivative of denominator is $2x$)$\int \frac{dx}{x^2+1} = \tan^{-1}x$Hence the original integral equals:$\frac{1}{5}\ln|x+2| - \frac{1}{10}\ln(x^2+1) + \frac{2}{5}\tan^{-1}x + c$.So $p = \frac{1}{5}$, $q = -\frac{1}{10}$, $r = \frac{2}{5}$.Therefore $p+q+r = \frac{1}{5} - \frac{1}{10} + \frac{2}{5} = \frac{2}{10} - \frac{1}{10} + \frac{4}{10} = \frac{5}{10} = \frac{1}{2}$.Answer:$\frac{1}{2}$ (Option B).