Blog: `(4x^2 - 1)/(2x(x + 1)^2)` Write the partial fraction decomposition of the rational expression.

Saturday, May 14, 2016

$\displaystyle\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

$\displaystyle\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}$

$\displaystyle\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}=\frac{{A}}{{{2}{x}}}+\frac{{B}}{{{x}+{1}}}+\frac{{C}}{{{\left({x}+{1}\right)}}^{{2}}}$

$\displaystyle\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}=\frac{{{A}{{\left({x}+{1}\right)}}^{{2}}+{B}{\left({2}{x}\right)}{\left({x}+{1}\right)}+{C}{\left({2}{x}\right)}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}$

$\displaystyle\therefore{\left({4}{{x}}^{{2}}-{1}\right)}={A}{{\left({x}+{1}\right)}}^{{2}}+{B}{\left({2}{x}\right)}{\left({x}+{1}\right)}+{C}{\left({2}{x}\right)}$

$\displaystyle{4}{{x}}^{{2}}-{1}={A}{\left({{x}}^{{2}}+{2}{x}+{1}\right)}+{B}{\left({2}{{x}}^{{2}}+{2}{x}\right)}+{2}{C}{x}$

$\displaystyle{4}{{x}}^{{2}}-{1}={A}{{x}}^{{2}}+{2}{A}{x}+{A}+{2}{B}{{x}}^{{2}}+{2}{B}{x}+{2}{C}{x}$

$\displaystyle{4}{{x}}^{{2}}-{1}={\left({A}+{2}{B}\right)}{{x}}^{{2}}+{\left({2}{A}+{2}{B}+{2}{C}\right)}{x}+{A}$

Therefore from the above,

$\displaystyle{A}+{2}{B}={4}$

$\displaystyle{2}{A}+{2}{B}+{2}{C}={0}$

$\displaystyle{A}=-{1}$

Solve the above equations for getting the values of A, B and C,

Substitute back the value of A in equation 1,

$\displaystyle-{1}+{2}{B}={4}$

$\displaystyle{2}{B}={4}+{1}={5}$

$\displaystyle{B}=\frac{{5}}{{2}}$

Substitute back the values of A and B in equation 2,

$\displaystyle{2}{\left(-{1}\right)}+{2}{\left(\frac{{5}}{{2}}\right)}+{2}{C}={0}$

$\displaystyle-{2}+{5}+{2}{C}={0}$

$\displaystyle{2}{C}+{3}={0}$

$\displaystyle{2}{C}=-{3}$

$\displaystyle{C}=-\frac{{3}}{{2}}$

$\displaystyle\therefore\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}=-\frac{{1}}{{{2}{x}}}+\frac{{5}}{{{2}{\left({x}+{1}\right)}}}-\frac{{3}}{{{2}{{\left({x}+{1}\right)}}^{{2}}}}$

$\displaystyle\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}={\left(\frac{{1}}{{2}}\right)}{\left(-\frac{{1}}{{x}}+\frac{{5}}{{{x}+{1}}}-\frac{{3}}{{{\left({x}+{1}\right)}}^{{2}}}\right)}$

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Saturday, May 14, 2016

$\displaystyle\frac{{{4}{{x}}^{{2}}-{1}}}{{{2}{x}{{\left({x}+{1}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

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