All Questions
Tagged with partial-fractions algebra-precalculus
157
questions
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Arbitrarily long decomposition into partial fractions.
$$\int \frac{dx}{x(x+p)(x+2p)(x+3p)...(x+(n-1)p)}= ?$$
I'm trying to solve this integral, and as I usually do in these cases, I break the expression into partial fractions, but I find this case ...
1
vote
1
answer
209
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Reasoning behind the resolution of partial fractions when denominator is the product of linear factors where some of them are repeating
The following text is from Mathematics for Class XII by Dr. R.D.Sharma, chapter "Indefinite Integrals", topic "Integration of Rational Algebraic Functions by using Partial Fractions&...
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2
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Partial fraction decomposition of $\frac{1}{x^a(x+c)^b}$
We have the partial fraction decomposition
$$\frac{1}{x^a(x+c)^b}=\frac{d_a}{x^a}+\frac{d_{a-1}}{x^{a-1}}+...+\frac{d_{1}}{x}+\frac{e_b}{(x+c)^b}+\frac{e_{b-1}}{(x+c)^{b-1}}+...+\frac{e_{1}}{x+c},$$
...
- 351
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2
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61
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Integration of rational of polynomials
I want to evaluate the indefinite integral for:
$$
\int\frac{x^3+3x−2}{x^2-3x+2}dx,\quad \text{for } x>2
$$
I did long division and factoring, simplifying it to
$$
\int x+3\,dx + \int\frac{10x-8}{(...
- 115
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1
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53
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How does this imply $x = z^2$?
I read that given the equation:
$$
x + \frac{1}{x} = z^2 + \frac{1}{z^2}
$$
, we can imply that $x=z^2$.
But that $x=z^2$ is not obvious to me... I know we can match the variables on the left hand ...
- 735
0
votes
1
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94
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Partial fractions Integration - Distributing Coefficients
Given the following Integral
$\int \frac{2x^3+ 2x^2+ 2x+ 1}{x^2 (x^2+1)}$
I would expand my fractions like the following
$\frac{A}{x} + \frac{B}{x^2}+\frac{Cx+D}{x^2+1}$
When I look at the ...
- 3
2
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4
answers
90
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How to get the value of $A + B ?$
I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(...
- 3,173
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1
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200
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Resolve into partial fractions $(x^2 + 3x - 5)/[(2x - 7)(x^2 + 3)^2]$
Resolve into partial fractions $\frac{x^2 + 3x - 5}{(2x - 7)(x^2 + 3)^3}$
The question has to do with the denominator being one linear and a repeated quadratic factor. Although, I am familiar with ...
1
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3
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253
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Integrating quadratics in denominator
I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
...
- 2,493
0
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3
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51
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Partial fraction expansion inquiry
How can I expand $\frac{a + 5}{(a^2-1)(a+2)}$ so that the sum of partial fractions is equal to $\frac{1}{a-1} - \frac{2}{a+1} + \frac{1}{a+2}$ ?
Thanks in advance!
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-1
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2
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Equations with 3 unknowns [closed]
I have this equation/problem:
Find the value of A and B
if 2/(x−5)(x+3) = A /(x−5) + B/(x+3)
How can I solve/approach this?
Thank you for your advice.
Regards
Lisa
- 43
2
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6
answers
222
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Why partial fraction decomposition of $\frac{1}{s^2(s+2)}$ is $\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$?
Can someone please explain why: $$\frac{1}{s^2(s+2)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$$
And not:$$\frac{1}{s^2(s+2)}=\frac{A}{s^2}+\frac{B}{(s+2)}$$
I'm a bit confused where the extra s ...
- 533
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1
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41
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How would I apply partial fraction expansion to this expression?
$$\displaystyle\frac{1}{X\bigg(1-\dfrac{X}{Y}\bigg)\bigg(\dfrac{X}{Z}-1\bigg)}$$
I want it in the form $$\frac{A}{X} + \frac{B}{(1-\dfrac{X}{Y})}+\frac{C}{(\dfrac{X}{Z}-1)}$$where I am required to ...
0
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1
answer
912
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Confusion about "picking values of $x$", partial fraction decomposition
One of the methods of decomposing a fraction and finding the constants, $A, B, C,$ etc., is to "pick values of $x$". For example, to find $A$ and $B$ of $${{3x}\over(x-1)(x+2)} = {A\over(x-1)} + {B\...
- 945
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0
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70
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Mistake in the computation via partial fractions
This is a computation made in Titchmash's Introduction to Zeta functions. I was trying to reverse the computation. However, I kept missing factors.
Consider $\frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}...
- 8,550