Mathematics Dictionary Dr. K. G. Shih

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Partial Fractions

Subjects

Read Symbol defintion Q01 | - Introduction Q02 | - Rules Q03 | - Find partial fraction of 1/(x^2-6*x+8) Q04 | - Find partial fraction of 1/(1-x^3) Q05 | - Q06 | - Q07 | - Q08 | - Q09 | - Q10 | -

Answers

Q01. Introduction Why do we need partial fractions ? One reason is to find integral. Eample : Find ∫dx/(x^2-6*x+8) Sine x^2 - 6*x + 8 = (x-2)*(x-4) Hence 1/(x^2 - 6*x + 8) = A/(x-2) + B/(x-4) Then we can find the integral ∫dx/(x^2-6*x+8) = ∫(A*dx)/(x-2) + ∫(A*dx)/(x-4) Go to Begin Q02. Rules The numerator in partial fraction is less one power than denominator Example 1 : (A*x + B)/(a*x^2 + b*x + c) Example 2 : (A*x^2 + B*x + C)/(a*x^3 + b*x^2 + c*x + d) Go to Begin Q03. Example : Find partial fraction of 1/(x^2 - 6*x + 8) Since x^2 - 6*x + 8 = (x-2)*(x-4) Hence 1/(x^2 - 6*x + 8) = A/(x-2) + B/(x-4) Both sides times (x^2 - 6*x + 8) 1 = A*(x-4) + B*(x-2) 1 = (A+B)*x - (4*A+2*B) Hence A+B = 0 ............. (1) 4*A + 2*B =-1 ............. (2) Solve (1) and (2) we have A = -1/2 and B = 1/2 Hence 1/(x^2 - 6*x + 8) = -1/(2*(x-2)) + 1/(2*(x-4)) Go to Begin Q04. Find partial fraction of 1/(1-x^3) Since 1 - x^3 = (1-x)*(1+x+x^2) Hence 1/(1-x^3) = A/(1-x) + (B*x+C)/(1+x+x^2) Both sides times (1-x^3) 1 = A*(1+x+x^2) + (B*x+C)*(1-x) 1 = (A+C) + (A+B-C)*x + (A-B)*x^2 Hence A+C = 1 ....................... (1) A+B-C = 0 ........................... (2) A-B = 0 ............................. (3) (2)+(3) we have 2*A-C = 0 ........... (4) (1)+(4) we have 3*A = 1 and A = 1/3 Hence B = A = 1/3 Hence C = 1-A = 2/3 Hence 1/(1-x^3) = 1/(3*(1-x)) + (1*x+2)/(1+x+x^2) Go to Begin Q05. Answer Go to Begin Q06. Answer Go to Begin Q07. Answer Go to Begin Q08. Answer Go to Begin Q09. Answer Go to Begin Q10. Answer Go to Begin

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