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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Simple partial fractions questions with fractions in the numerator.

$$ \dfrac{1}{\left(x+1\right)\left(2x+1\right)}$$ Using the cover up rule, I received the answer of $$\dfrac{\dfrac{1}{2}}{2x+1}-\frac{1}{x+1}$$ or $$\dfrac{1}{2(2x+1)}-\frac{1}{x+1}$$ Answer at the back of the book…
Modrisco
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Partial Fraction Expansion of Transfer Function

How do I go from: $$ \frac{3(1+0.2z^{-1})(1+z^{-1})}{(1+0.5z^{-1})(1-0.4z^{-1})} $$ to $$ -3 + \frac{7}{1-0.4z^{-1}} - \frac{1}{1+0.5z^{-1}} $$ I understand that the first form can be expanded as $$ \frac{A_1}{1-0.4z^{-1}} +…
Aeon
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Partial fraction decomposition of $\frac{x-1}{x^3+x^2}$

Why the partial fraction expansion of $\frac{x-1}{x^3+x^2}$ is $$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$$ if I only have that $x^3 + x^2$ is the same as $x^2(x + 1)$?
juliano.net
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partial fractions specific question

so i thought i knew about partial fractions, but apparently i don't. i have the answer to a partial fraction but i can't figure out how you get to that answer. the value is $$ X(z) = Z*(Z+2)/(Z-1)^2 = 1 + 3z/(z-1)^2+1/(z-1) $$ Where does the "1"…
pato.llaguno
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Decomposing partial fraction with a denominator negative quadratic expression $-x^2$

${5x+4\over -x^2-x+2}$ My solution: ${5x+4\over -x^2-x+2}$ $= {5x+4\over -(x-1)(x+2)}$ $= {A\over (x-1)} + {B\over (x+2)}$ $= {3\over -(x-1)} + {2\over -(x+2)}$ $= -{3\over x-1} -{2\over x+2}$ Can I multiply both fractions by -1 as follows: Rather…
direprobs
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Finding the partial fraction of $x+21/2(2x+3)(3x-2)$

$x+21 / 2(2x+3)(3x-2)$ The 2 in the denominator represents a problem for me: $x+21 / 2(2x+3)(3x-2) = A / 2(2x+3) * B / 2(3x-2)$ $x+21 = A(2)(3x-2) + B(2x+3)$
direprobs
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Expanding a partial fraction...

$\dfrac1{x^2+3}=x^{-2}\left(1+\dfrac{3}{x^2}\right)^{-1} = x^{-2}\left(1-\dfrac{3}{x^2}+\dfrac{9}{x^4}+\cdots\right)$ But this can also be factored into: $\dfrac{1}{3}\left(1+\dfrac{x^2}{3}\right)^{-1} =…
user61871
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What formula do I use when i have to find the partial fraction of

What formula do I use when i have to find the partial fraction of $$\frac{10x^2+11x+19 }{ (x-0.5)(2x^2+6x+10)}$$ Is it $A(2x^2+6x+10) + (Bx+C)(x-0.5)$ ? Or do I have to factorise $(2x^2+6x+10)$ into $2(x^2+6x+10)$ and then use another formula?
user164612
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Why is that integral written as the product of 2 other integrals?

I am a student and just learning about integrals. **Could you please explain to me how the highlighted part of the equation was derived?** I can't quite understand how was the c/a term brought of the integral and how 1 integral just became a…
Butyl
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Partial Fraction Decomposition for Inverse Laplace Transform

I'm using John Bird's Higher Engineering Mathematics but I'm really struggling with Partial Fraction decomposition. I'm needing to convert it from s domain to t domain using Inverse Laplace Transform. Would really appreciate a bit of guidance, not…
Loz Marks
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how to partially decompose this fraction?

I have: $ \frac{d\tau}{d Z}=\frac{N}{(P-Z)Z} $ And apparently this function can be rewritten as: $ \frac{d\tau}{d Z}=\frac NP \left(\frac1{P-Z}+\frac 1Z\right) $ for further integration. But how must I rewrite this function? I…
Gerard de Jong
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How to prove the given equation?

While studying Ramanujan's Notebooks I stumbled upon the following entry : $\frac{1}{(x^3-x)} = \frac{1}{2(x+1)} + \frac{1}{2(x-1)} - \frac{1}{x}$ Now, I can prove this identity while working on the RHS. However, if I were to prove it by expanding…
tanya
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A Partial fraction expansion questions.

I am learning signals and systems. I solve the problem and reach the equation (5.164). How can I work out the value of $A$ and $B$? I do the partial fraction expansion and yields
lzane
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Partial fraction of non repeating quadratic factors

$\frac{ 2(s^2 + 9(s-1)}{ s (s^2 -9) } = \frac{A}{S} + \frac{ Bs + C}{s^2 -9} $ $ 2(s^2 + 9(s-1) ) = A (s^2 -9) + Bs+C $ Let $S = 0$ $ 2 (-18) = A(-9)$ , so $A = 2$ I sub $A = 2$ into the equation to get $2s^2 + 18s - 18 = s^2 (2 + B) + Cs - 18$…
user185692
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decomposition into partial fractions

could you please help me with following partial fractions ? 1) $$ \frac{1}{x(x^2+1)} $$ 2) $$ \frac{1}{1+x^3} $$ 3) $$ \frac{x}{x^3-1} $$ Thank you for very much for your time
Radoslav Ani Smolík
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