How do you write the partial fraction decomposition of rational expression?
Hello! I m having a bit of trouble with this problem: 1/x^2+x I factored the denominator, to be: 1/x (x+1) I wrote it in decomposition form: A/x + B/x+1 I wrote it as a basic equation: A(x+1)/x(x+1) + Bx/x(x+1) 1 = A(x+1) + Bx And distributed: 1=Ax+A+Bx From there I got stuck. I looked in the textedbook and my class notes, but I m still having a hard time finding the partial fraction decomposition. Did I do something wrong that made me get stuck? I thought, by looking at the 1, I would write 1=A + B, because there s one x for A and B. Though, that made me doubt too. Will you help me? Please? Thank you for your time!
Help finding the partial fraction decomposition.?
The above answers incorrectly interpreted your question, as division comes before addition by default. There is no possible partial fraction expansion for this expression.
What did I do wrong in this partial fraction decomposition problem (picture linked)?
-2a+4b-14 =0The mistake is that you put after 2a+4b -14 ….2a is the mistake should be -2a .So a+b=8 and -2a+4b -14=0a=8-b and -2a+4b=14=> -2(8-b) +4b =14=> -16+2b+4b=14=>6b=16+14=30 => b=30/6 =5=> a=8-b = 8–5=3=>[math]\frac{8x+14}{(x+4)(x-2)} = \frac{3}{x+4} + \frac{5}{x-2}[/math]
Evaluate the integral using Partial Fractions Decomposition (Calc2)?
first divide...then you will have something like x + linear / { [x+4] [x+3] } , do partial on the fraction linear / { [..][..] } = a / [x+4] + b / [ x + 3]...multiply both sides by x + 4 and then let x = -4 to find ' a ' , by x + 3 and let x = -3 to find ' b '...then integrate...try it
Integral of (x-7)/(x^2-x-12) using partial fraction decomposition?
(x-7)/(x^2-x-12) (x-7)/(x-4)(x+3) (x-7)/(x-4)(x+3) = A/(x-4) + B/(x+3) x - 7 = A(x+3) + B(x-4) let x = 4 -3 = 7A A = -3/7 let x = -3 -10 = -7B B = 10/7 int : (x-7)/(x^2-x-12) int : (-3/7)/(x-4) + (10/7)/(x+3) (10/7)ln|x+3| - (3/7)ln|x-4| + C I got the same answer. I am not sure why it is wrong. What is the right answer? do you know?
Does it matter when using partial fraction decomposition the order I choose?
Hello, it doesn't matter at all; let's see the decomposition: 1 /(x^4 - x²) = 1 /[x² (x² - 1)] = 1 /[x² (x + 1)(x - 1)] = A/x + B/x² + C/(x + 1) + D/(x - 1) 1 /[x² (x + 1)(x - 1)] = [Ax (x + 1)(x - 1) + B(x + 1)(x - 1) + Cx² (x - 1) + Dx² (x + 1)] /[x² (x + 1)(x - 1)] (equating numerators) 1 = Ax (x² - 1) + B(x² - 1) + Cx² (x - 1) + Dx² (x + 1) 1 = Ax³ - Ax + Bx² - B + Cx³ - Cx² + Dx³ + Dx² 1 = (A + C + D)x³ + (B - C + D)x² + (- A)x + (- B) (equating coefficients) A + C + D = 0 B - C + D = 0 - A = 0 - B = 1 0 + C + D = 0 - 1 - C + D = 0 A = 0 B = - 1 C = - D - 1 - (- D) + D = 0 → - 1 + D + D = 0 → 2D = 1 A = 0 B = - 1 C = - D = - 1/2 D = 1/2 A = 0 B = - 1 yielding: 1 /[x² (x + 1)(x - 1)] = A/x + B/x² + C/(x + 1) + D/(x - 1) = - 1/x² + (-1/2)/(x + 1) + (1/2)/(x - 1) thus the integral becomes: ∫ {- (1 /x²) - [(1/2) /(x + 1)] + [(1/2) /(x - 1)]} dx = (splitting and pulling constants out) - ∫ x^(- 2) dx - (1/2) ∫ [1 /(x + 1)] dx + (1/2) ∫ [1 /(x - 1)] dx = - [1/(- 2+1)] x^(- 2+1) - (1/2) ln |x + 1| + (1/2) ln |x - 1| + C = - [1/(- 1)] x^(- 1) + (1/2) (ln |x - 1| - ln |x + 1|) + C = (applying logarithm properties) (1 /x) + (1/2) ln |(x - 1) /(x + 1)| + C as a matter of fact I get the same answer as yours; then, since I have just checked it successfully, evidently your book is wrong... I hope it helps
How do I decompose [math] \frac {x^3-2x^2+4x-2}{x^4-2x^3+2x^2}[/math] using partial fractions ?
Factor [math]x^2[/math] from the denominator:[math]\dfrac{x^3–2x^2+4x-2}{x^4–2x^3+2x^2}=\dfrac{x^3–2x^2+4x-2}{x^2(x^2–2x+2)}[/math]The factors in the denominator are at their most simplified form. Express the function as the sum of partial fractions (both factors have degree 2):[math]\begin{aligned}\dfrac{x^3–2x^2+4x-2}{x^2(x^2–2x+2)}&=\dfrac{Ax+B}{x^2}+\dfrac{Cx+D}{x^2–2x+2}\\x^3–2x^2+4x-2&=(Ax+B)(x^2–2x+2)+Cx^3+Dx^2\\&=Ax(x^2–2x+2)+B(x^2–2x+2)+Cx^3+Dx^2\end{aligned}[/math]If [math]x=0,[/math] then the terms in [math]A, C, D[/math] cancel out. Therefore:[math]0^3–2\cdot 0^2+4\cdot 0-2=2B\implies B=-1[/math]Plugging the value of [math]B[/math] back in the equation and grouping by power of [math]x[/math]:[math]\begin{aligned}x^3–2x^2+4x-2&=Ax(x^2–2x+2)-(x^2–2x+2)+Cx^3+Dx^2\\x^3-x^2+2x&=Ax(x^2–2x+2)+Cx^3+Dx^2\\&=Ax^3–2Ax^2+2Ax+Cx^3+Dx^2\\&=x^3(A+C)-x^2(2A-D)+2Ax\end{aligned}[/math]This in turn gives us a system of equations with the coefficients of [math]x[/math]:[math]\begin{cases} A+C=1 \\ 2A-D=1 \\ 2A=2 \end{cases}[/math]Which implies that [math]A=1, C=0, D=1[/math]. Plugging back those values:[math]\begin{aligned}\dfrac{x^3–2x^2+4x-2}{x^4–2x^3+2x^2}&=\dfrac{Ax+B}{x^2}+\dfrac{Cx+D}{x^2–2x+2}\\&=\dfrac{x-1}{x^2}+\dfrac{1}{x^2–2x+2}\\&=\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^2–2x+2}\end{aligned}[/math]
How do you decompose a fraction?
I infer that the questioner is not mathematically sophisticated, as he has brackets in the wrong places. The question is equivalent to:"Find a and b such that this equation is an identity, that is, true for all n."In that case it becomes the simplest and most common partial fraction decomposition, and a solution method is demonstrated by Carter McClung.
Why do partial fractions work?
I will assume you mean partial fraction decomposition. Basically, it is the reverse step of what you do when you consolidate multiple fractions. However, the difference is in the application of the fundamental theorem of algebra to this end. Because we can take a sum or difference of multiple fractions and take them under one division symbol, we should be able to decompose a single complicated fraction into a sum or difference of multiple fractions. There exist cases where this is simply impossible, and the fundamental theorem of algebra allows us to predict and define these cases.see: irreducible polynomial factor