Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [partial-fractions]

Rewriting rational function in the form of partial fractions is often useful when calculating integrals.

Rewriting rational function in the form of partial fractions is often useful when calculating integrals. The possibility of decomposing a rational function into a sum of simplified fractions is guaranteed by the fundamental theorem of algebra.

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Partial fractions integration

Re-express $\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}$ in terms of partial fractions and compute the indefinite integral $\dfrac{1}5{}\int f(x)dx $ using the result from the first part of the question.
neverloggedin
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Partial fraction decomposition of $\frac{1}{x^4-x^2}$

I have been having a debate over whether, when you factor the denominator into $x^2$, $x-1$, and $x+1$, you need a fraction that says $\frac{A}{x}$ and one that has $\frac{B}{x^2}$ or if you only need the fraction with $x^2$ as the denominator. When…
mmm
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partial fractions decomposition issue

How do you decompose the following fraction? $$\frac{2}{2x^2 (x+1)} = \frac{a}{2x^2} + \frac{b}{x+1}$$ i just want to know if I am on the right track or I am missing something(based on the above beginning).
Lucas
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How to split complex fractions like $\frac{e^{-as}}{s(1-e^{-as})}$ using partial fraction technique?

$$\frac{e^{-as}}{s(1-e^{-as})}=\frac{A}{s}+\frac{B}{1-e^{-as}}$$ Multiply the whole thing by $s(1-e^{-as})$ $$e^{-as}=A(1-e^{-as})+Bs$$ I distribute and try to find A and B by matching coefficients. I get $A=1, -1$ and $B=0$. Now what? Turns out by…
most venerable sir
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How do I express this expression using partial fractions?

The expression I'm trying to express using partial fractions is: $\frac{64x}{(2x-1)^2(2x+1)^2}$ What I've tried so far: $\frac{64x}{(2x-1)^2(2x+1)^2}=\frac{A}{(2x-1)^2}+\frac{B}{2x-1}+\frac{C}{(2x+1)^2}+\frac{D}{2x+1}$ where $A,B,C,D$ are…
thebuddha
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given an equation, find A and B

I can easily solve this problem by finding A and B, and then A+B. My question is where there is a way to obtain A+B without finding A and B first. The problem is supposed to be challenging, but it looks too easy. That is why I think there must be a…
learning
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Question about partial fractions and the order of two linear factors

I have a question on the topic of partial fraction and decomposition. Lets say I have the integral of $\frac{1}{(x^2 - x - 2)}$. When I want to find out the A and B values, I first have to get linear or quadratic factors (in this case linear). So I…
Paul Lee
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How to Decompose into Partial Fractions?

I need to decompose (1/((s - 2)^(2) + 1^(2))) into partial fractions, but I am not sure how exactly. Here are my attempts: Attempt 1 (1/((s - 2)^(2) + 1^(2))) = (A/(s - 2)) + (B/(s - 2)^(2)) + (C/(1)) + (D/(1^(2))) = (A/(s - 2)) + (B/(s - 2)^(2)) +…
Khaltazar
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Proof using partial fractions

I have to prove this formula: $$\int \! \frac{1}{(x^2+\beta ^2)^{k+1}} \, \mathrm{d}x=\frac{1}{2k\beta ^2}\frac{x}{(x^2 +\beta^2)^k}+\frac{2k-1}{2k\beta^2}\int \! \frac{1}{(x^2+\beta ^2)^k} \, \mathrm{d}x $$ I have to use partial fraction…
Arthur
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Partial Fractions- Is there a quicker way?

So with the fraction $$ (2s^2+5s+7)/(s+1)^3 $$ Is there a quicker way to solve this rather than equating the coefficients? I cant use the 'cover-up' method because its just one fraction This is what I got: $$ A/(s+1) + B/(s+1)^2 + C/(s+1)^3 $$
Navy Seal
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Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$

The partial fraction of $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$. I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?
Stanko
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Partial Fraction Decomposition Problem

I am having trouble with this problem. I need to integrate: $$\frac1{T^4}\times \frac1{K-T}$$ with respect to $T$. If I do PFD: $$\frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T}$$ Does this look right? From here, I am…
Jackson Hart
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Partial fraction decomposition of type $1/(x^2+k)$

I know that partial fraction of this can be written as: $$\frac{3x}{(1+x)(2+x)}=\frac{-3}{1+x}+\frac{6}{2+x}$$ Which can be done in these ways: $$\frac{3x}{(1+x)(2+x)}=\frac{A}{1+x}+\frac{B}{2+x}\implies3x=A(2+x)+B(1+x),\forall\;x$$ And now solving…
RE60K
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Partial fraction (doubt)

I have this partial fraction $$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$ I tried to resolve using this…
LP0
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Help w/ Partial Fraction Decomposition

I need some help figuring out how to decompose $\displaystyle\frac{1}{x^4+1}$ into partial fractions. This is what I have done so far: $$\frac{1}{x^4+1} = \frac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}$$ From there, I do not know how to…
A is for Ambition
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