Partial Fraction Decomposition

Solved splits with coefficient comparison • SaitechAI
(10)
\[ \frac{x+2}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+1} \]
Clear denominators and equate coefficients.
\[ x+2 = A(x^2+1) + (Bx+C)(x+1) \] \[ x+2 = (A+B)x^2 + (B+C)x + (A+C) \] \[ \begin{cases} A + B = 0,\\ B + C = 1,\\ A + C = 2 \end{cases} \Longrightarrow A=\tfrac12,\; B=-\tfrac12,\; C=\tfrac32 \]
\[ \boxed{\frac{x+2}{(x+1)(x^2+1)} = \frac{\tfrac12}{x+1} + \frac{-\tfrac12 x + \tfrac32}{x^2+1}} \]
Split: simple + irreducible quadratic
(11)
\[ \frac{7x^2-25x+6}{(x^2-2x-1)(3x-2)} = \frac{Ax+B}{x^2-2x-1} + \frac{C}{3x-2} \]
Clear denominators, expand, and compare coefficients.
\[ 7x^2-25x+6 = (Ax+B)(3x-2) + C(x^2-2x-1) \] \[ 7x^2-25x+6 = (3A+C)x^2 + (-2A+3B-2C)x + (-2B-C) \] \[ \begin{cases} 3A + C = 7,\\ -2A + 3B - 2C = -25,\\ -2B - C = 6 \end{cases} \Longrightarrow A=1,\; B=-5,\; C=4 \]
\[ \boxed{\frac{7x^2-25x+6}{(x^2-2x-1)(3x-2)} = \frac{x-5}{x^2-2x-1} + \frac{4}{3x-2}} \]
Split: quadratic factor + linear factor
(12)
\[ \frac{x^2+x+1}{x^2+2x+1} = A + \frac{B}{x+1} + \frac{C}{(x+1)^2} \quad(\text{since }x^2+2x+1=(x+1)^2) \]
Clear denominators and match coefficients.
\[ x^2+x+1 = A(x+1)^2 + B(x+1) + C \] \[ x^2+x+1 = A x^2 + (2A+B)x + (A+B+C) \] \[ \begin{cases} A=1,\\ 2A+B=1,\\ A+B+C=1 \end{cases} \Longrightarrow A=1,\; B=-1,\; C=1 \]
\[ \boxed{\frac{x^2+x+1}{(x+1)^2} = 1 - \frac{1}{x+1} + \frac{1}{(x+1)^2}} \]
Split: repeated linear factor