Blog: `(x + 1)/(x^3(x^2 + 1)^2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Wednesday, June 19, 2013

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{3}}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

You need to decompose the fraction in simple irreducible fractions, such that:

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{3}}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}=\frac{{A}}{{x}}+\frac{{B}}{{{{x}}^{{2}}}}+\frac{{C}}{{{{x}}^{{3}}}}+\frac{{{D}{x}+{E}}}{{{{x}}^{{2}}+{1}}}+\frac{{{F}{x}+{G}}}{{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}$

You need to bring to the same denominator all fractions, such that:

$\displaystyle{x}+{1}={A}{\left({{x}}^{{2}}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}\right)}+{B}{x}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}+{C}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}+{\left({D}{x}+{E}\right)}{{x}}^{{3}}\cdot{\left({{x}}^{{2}}+{1}\right)}+{\left({F}{x}+{G}\right)}\cdot{{x}}^{{3}}$

$\displaystyle{x}+{1}={A}{{x}}^{{2}}{\left({{x}}^{{4}}+{2}{{x}}^{{2}}+{1}\right)}+{B}{x}{\left({{x}}^{{4}}+{2}{{x}}^{{2}}+{1}\right)}+{C}{\left({{x}}^{{4}}+{2}{{x}}^{{2}}+{1}\right)}+{\left({D}{{x}}^{{4}}+{E}{{x}}^{{3}}\right)}\cdot{\left({{x}}^{{2}}+{1}\right)}+{F}{{x}}^{{4}}+{G}{{x}}^{{3}}$

$\displaystyle{x}+{1}={A}{{x}}^{{6}}++{2}{A}{{x}}^{{4}}+{A}{{x}}^{{2}}+{B}{{x}}^{{5}}+{2}{B}{{x}}^{{3}}+{B}{x}+{C}{{x}}^{{4}}+{2}{C}{{x}}^{{2}}+{C}+{D}{{x}}^{{6}}+{D}{{x}}^{{4}}+{E}{{x}}^{{5}}+{E}{{x}}^{{3}}+{F}{{x}}^{{4}}+{G}{{x}}^{{3}}$

You need to group the terms having the same power of x:

$\displaystyle{x}+{1}={{x}}^{{6}}{\left({A}+{D}\right)}+{{x}}^{{5}}{\left({B}+{E}\right)}+{{x}}^{{4}}{\left({2}{A}+{C}+{D}+{F}\right)}+{{x}}^{{3}}{\left({2}{B}+{E}+{G}\right)}+{{x}}^{{2}}{\left({A}+{2}{C}\right)}+{x}{\left({B}\right)}+{C}$

Comparing the expressions both sides yields:

$\displaystyle{A}+{D}={0}$

$\displaystyle{B}+{E}={0}$

$\displaystyle{2}{A}+{C}+{D}+{F}={0}\Rightarrow{F}={1}$

$\displaystyle{2}{B}+{E}+{G}={0}\Rightarrow{G}=-{1}$

$\displaystyle{A}+{2}{C}={0}\Rightarrow{A}=-{2}\Rightarrow{D}={2}$

$\displaystyle{B}={1}\Rightarrow{E}=-{1}$

$\displaystyle{C}={1}$

Hence, the partial fraction decomposition of the improper rational expression is $\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{3}}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}=-\frac{{2}}{{x}}+\frac{{1}}{{{{x}}^{{2}}}}+\frac{{1}}{{{{x}}^{{3}}}}+\frac{{{2}{x}-{1}}}{{{{x}}^{{2}}+{1}}}+\frac{{{x}-{1}}}{{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}$

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Wednesday, June 19, 2013

$\displaystyle\frac{{{x}+{1}}}{{{{x}}^{{3}}{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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