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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Strange Partial Fractions Decomposition

I am trying to get from $$\frac{z^7 + 1}{z^2(z^4+1)}$$ to $$\frac{1}{z^2} + z - \frac{z+z^2}{1+z^4}.$$ The author did this by doing a partial fractions decomposition. I don't see how, however.. If I compute the partial fractions decomposition, I…
Pascal Engeler
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What does it mean to equate the coefficients of like terms when solving for A and B in partial fractions?

I'm trying to step myself through solving partial fractions in a year 10 book by Cambridge. This is a concept they're introducing early for students who want to challenge themselves and it's pretty light on the explanation. For example: 7/( x+2 ) (…
duckegg
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Find partial fractions of $\frac{3x^2-2x+1}{x^2(1-x^2)}$ and then deduce partial fractions of $\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$

Find partial fractions of, $$\frac{3x^2-2x+1}{x^2(1-x^2)}$$ Hence deduce partial fractions of $$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$$ My Try I was able to do the first…
emil
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Partial fraction decomposition of fraction with non-polynomial terms

If I for example have a fraction \begin{align} \frac{1}{(x-e) \space \ln (x)} \end{align} can I use partial fraction decomposition to get an equation like this? \begin{align} \frac{A}{x-e} + \frac{B}{\ln (x) }\end{align} When I tried to do the…
eirik-ff
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Why do you need two fractions for partial fraction decomposition with repeated factors?

For example, suppose my denominator contains $(x - 1)(x - 1)$. I know I need two fractions, one with $(x - 1)$ and one with $(x - 1)^2$ as the denominator. But I'm looking for a deeper reason as to why. It makes sense when you go through and get a…
mmm
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Partial fraction: is my final answer and computation correct?

I need some help at this exercise (partial fraction): $$\int{\frac{4x + 1}{x^3(x+2)}} dx$$ First of all I calculated the roots of the denominator. $$ x_{1, 2, 3} = 0 \\ x_4 = -2 $$ After that I assigned the roots to the partial…
user23053
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Integration of fractions and cancellation

How come $$\int\frac{3}{6+3x}\mathrm{d}x=\mathrm{ln}(2+x)+C$$ and not $\ \mathrm{ln}(6+3x)+C$ ? I understand you can cancel the fraction but why should this make a difference? I've tried searching through all my notes on this and online but the…
user245752
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Getting Partial Fraction Decomposition Wrong

So I have done this solution: But it's wrong according to technology. I fail to see any error I've made in this solution, do you guys have any idea what's wrong?
KlingL
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Values tried for partial fraction decomposition

I'll explain my question with the following example from wikipedia. Suppose, we have a function: $$ f(x)=\frac{1}{x^2+2x-3} $$ Here, the denominator splits into two distinct linear factors: $$ q(x)=x^2+2x-3 = (x+3)(x-1) $$ so we have the partial…
shobhu
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Problem with partial fractions

I have this integral: \begin{equation*} \int \! \frac{x-1}{(x+1)(x^2+9)} \, \mathrm{d}x. \end{equation*} I already split the denominator into two factors. Now, when I do partial fraction decomposition, I have this equation: $x-1=a(x^2+9)+b(x+1)$.…
Arthur
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How to Decompose into Partial Fractions

Why is it useful to write: $$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$ and not: $$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$ when decomposing into partial fractions?
mrk
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Need help with converting $\frac{1}{(x-1)(x+2)^2}$ to partial fractions

So far I have my working ... I think I got wrong ... since after checking with http://calc101.com/webMathematica/partial-fractions.jsp#topdoit $A = \frac{1}{9}$ $B = - \frac{1}{9}$ $C = - \frac{1}{3}$ Is there a better way to do this than…
Jiew Meng
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Partial Fractions of form $\frac{1}{(ax+b)(cx+d)^2}$

When asked to convert something like $\frac{1}{(ax+b)(cx+d)}$ to partial fractions, I can say $$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$ Then why can't I split $(cx+d)^2$ into $(cx+d)(cx+d)$ then do $$\frac{1}{(ax+b)(cx+d)^2} =…
Jiew Meng
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How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$?

Please could anyone help me to integrate $\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$ I know how to use partial fraction and I did this: $$ x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right) $$ And then ?.$\quad$ Thanks all.
user2849967
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Determining which values to use in place of x in functions

When solving partial fractions for integrations, solving x for two terms usually isn't all that difficult, but I've been running into problems with three term integration. For example, given $$\int\frac{x^2+3x-4}{x^3-4x^2+4x}$$ The denominator…
Jason
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