How many cases are there in integration using partial fractions?
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I'm not quite sure what you mean by your question - could you explain a bit more what you mean by a "case"? – Mark Bennet Apr 24 '14 at 14:12
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I appreciate you trying to help Sir @MarkBennet. Thanks – anakin skywalker Apr 24 '14 at 14:27
1 Answers
If I understood your question correctly, I would say there are $5$ cases.
Assume you have a rational function $\dfrac{p(x)}{q(x)}$, where the degree of $q(x)$ exceeds the degree of $p(x)$.
Case $1$: $q(x)$ is a product of distinct linear factors
Example: Consider $q(x)=\dfrac{x}{(x+3)(x-1)}$
Case $2$: $q(x)$ is a product of linear factors, where some of these factors are repeated
Example: Consider $q(x)=\dfrac{x^2}{(x+4)^2(x-2)}$
Case $3$: $q(x)$ is a product of distinct irreducible quadratic factors
Example: Consider $q(x)=\dfrac{x}{(x^2+1)(x^2+3)}$
Case $4$: $q(x)$ is a product of irreducible quadratic factors, where some are repeated
Example: Consider $q(x)=\dfrac{2x-1}{(x^2+x+1)^3}$
Case $5$: $q(x)$ is some mixture of the above cases.
Example: Consider $\dfrac{3x-2}{(x-2)^2(x^2+x+2)}$
Consider another example, which has been worked to the decomposition stage of the solution.
\begin{align} &\frac{2x-1}{(x-1)^2(x^2+x+1)^2}\\ &=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+x+1}+\frac{Ex+F}{(x^2+x+1)^2}\\ \end{align}
Then, you can do what you normally do for partial fractions and equate the coefficients.
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Sir, can you give me an example of case number 5? @Sujaan Kunalan – anakin skywalker Apr 24 '14 at 14:24
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