All Questions
Tagged with partial-fractions algebra-precalculus
157
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About a statement of partial fraction in an answer
I'm reading this answer of The logic behind partial fraction decomposition, I think my question is too basic and not directly related to the answer so I don't comment there. I don't understand why:
...
- 229
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4
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Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$
I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just tried to work out the ...
- 29
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3
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How do we make $A(x+1)(x^2+4x+5)+B(x^2...$ to be equal to $2x^2$??
$$\frac{2x^2}{(x+1)^2(x^2+4x+5)}$$
$$2x^2=A(x+1)(x^2+4x+5)+B(x^2+4x+5)+(Cx+D)(x+1)^2$$
We can get $B=1$ if we put $-1$ for $x$.
But I don't know how can we solve for $A$, $C$ and $D$ since we can't ...
1
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3
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Partial Fraction problem solution deviates from the Rule
Question:
Compute $\displaystyle \int\frac{x^2+1}{(x^2+2)(x+1)} \, dx$
My Approach:
As per my knowledge this integral can be divided in partial Fraction of form $\dfrac{Ax+B}{x^2+px+q}$ and then do ...
user517784
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1
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Find the Partial Fraction Decomposition
$\frac{2x^5+3x^4-3x^3-2x^2+x}{2x^2+5x+2}$, I am not sure as to where to start with this one; I have already done the factoring process of the denominator but not sure how to continue the algebraic ...
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A quick way for decomposing fractions
The complete method for decomposing fractions is obvious . For example $y = \frac{2x+1}{(x-1)(x+3)} = \frac{a}{x-1} + \frac{b}{x+3} = \frac{a(x+3) + b(x-1)}{(x-1)(x+3)} \Rightarrow $$
\left\{
\begin{...
- 4,359
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5
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Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$
Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in ...
- 671
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partial fraction decomposition of $\frac{k^4}{(a\, k^3-1)^2}$
I have to perform complex partial fraction decomposition of the following term:
$$\frac{k^4}{(a \, k^3-1)^2}$$
where $a$ is a real positive number.
and I would like to know if it is possible to ...
- 563
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Why can I decompose fractions this way? [duplicate]
In order to be able to find an antiderivative, I have been taught to decompose fractions by first observing an identity, like $$\frac{2x+1}{(3x-2)(x+1)} \equiv \frac{A}{3x-2}+\frac{B}{x+1}$$
wherever ...
- 1,750
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4
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Having trouble using my usual method of partial fraction decomposition for $\frac{9 + 3s}{s^3 + 2s^2 - s - 2}$.
I'm having trouble using my usual method of partial fraction decomposition for $\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2}$.
We can factor such that $$\dfrac{9 + 3s}{s^3 + 2s^2 - s - 2} = \dfrac{A}{s - 1} + \...
- 4,322
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Suppose $a+b+c+d=s$ and $1/a+1/b+1/c+1/d=s$, how can you make any possible conclusions?
It is known that there is such a number $s$ with real numbers $a,b,c,d$ that are not equal to $0$ nor $1$ and satisfy the following equations.
$$ a+b+c+d=s$$
$$1/a+1/b+1/c+1/d=s$$
$$\frac{1}{1-a}+\...
- 146
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3
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Partial Fractions help!?
$A + C = 0$
$-4A + B - 8C + D = 1$
$3A + 16C - 8D = -29$
$-12A + 3B + 16D = 5$
How do I equate the coefficients? Please provide steps an an explanation.
- 33
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What are the criteria for partial fractions?
Consider the example of
$$1\over (u^2 - 1)(u+1)$$.
$${Au + B\over u^2 - 1} + {C \over u + 1} = {Au^2 + Au + Bu + B + Cu^2 - C\over (u^2 - 1)(u+1)} = {1\over (u^2 - 1)(u+1)}$$
$$\begin {cases}A + ...
user312097
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partial fractions for a function
I need help finding the partial fraction decomposition for this function, I am just lost on it, here it is:
$(x^2 + x + 1)/(2x^4+3x^2+1)$. the help is appreciated. thank you.
- 157
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Why does Partial Fraction Decomposition Result in Multiples of the Decomposed Fraction?
Let's say I have the rational function $\dfrac{x^3}{x^2 + x - 6}$. I use polynomial long division to get $x^3 = (x - 1)(x^2 + x - 6) + 7x -6 \implies \dfrac{x^3}{(x - 2)(x + 3)} = \dfrac{x^3}{x^2 + x -...
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