Notes: `int 2/(9x^2-1) dx` Use partial fractions to find the indefinite integral

Tuesday, 27 January 2015

$\displaystyle\int\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

$\displaystyle\int\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}$

To solve using partial fraction method, the denominator of the integrand should be factored.

$\displaystyle\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}=\frac{{2}}{{{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}}}$

Then, express it as sum of fractions.

$\displaystyle\frac{{2}}{{{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}}}=\frac{{A}}{{{3}{x}-{1}}}+\frac{{B}}{{{3}{x}+{1}}}$

To solve for the values of A and B, multiply both sides by the LCD of the fractions present.

$\displaystyle{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}\cdot\frac{{2}}{{{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}}}={\left(\frac{{A}}{{{3}{x}-{1}}}+\frac{{B}}{{{3}{x}+{1}}}\right)}\cdot{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}$

$\displaystyle{2}={A}{\left({3}{x}+{1}\right)}+{B}{\left({3}{x}-{1}\right)}$

Then, assign values to x in which either (3x+1) or (3x-1) will become zero.

So plug-in x=1/3 to get the value of A.

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$\displaystyle\int\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}$

To solve using partial fraction method, the denominator of the integrand should be factored.

$\displaystyle\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}=\frac{{2}}{{{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}}}$

Then, express it as sum of fractions.

$\displaystyle\frac{{2}}{{{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}}}=\frac{{A}}{{{3}{x}-{1}}}+\frac{{B}}{{{3}{x}+{1}}}$

To solve for the values of A and B, multiply both sides by the LCD of the fractions present.

$\displaystyle{2}={A}{\left({3}{x}+{1}\right)}+{B}{\left({3}{x}-{1}\right)}$

Then, assign values to x in which either (3x+1) or (3x-1) will become zero.

So plug-in x=1/3 to get the value of A.

$\displaystyle{2}={A}{\left({3}\cdot\frac{{1}}{{3}}+{1}\right)}+{B}{\left({3}\cdot\frac{{1}}{{3}}-{1}\right)}$

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

$\displaystyle{2}={A}{\left({2}\right)}+{B}{\left({0}\right)}$

$\displaystyle{2}={2}{A}$

$\displaystyle{1}={A}$

Also, plug-in $\displaystyle{x}=-\frac{{1}}{{3}}$ to get the value of B.

$\displaystyle{2}={A}{\left({3}\cdot{\left(-\frac{{1}}{{3}}\right)}+{1}\right)}+{B}{\left({3}\cdot{\left(-\frac{{1}}{{3}}\right)}-{1}\right)}$

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

$\displaystyle{2}={A}{\left({0}\right)}+{B}{\left(-{2}\right)}$

$\displaystyle{2}=-{2}{B}$

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

So the partial fraction decomposition of the integrand is

$\displaystyle\int\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left(\frac{{2}}{{{\left({3}{x}-{1}\right)}{\left({3}{x}+{1}\right)}}}{\left.{d}{x}\right.}\right.}$

$\displaystyle=\int{\left(\frac{{1}}{{{3}{x}-{1}}}-\frac{{1}}{{{3}{x}+{1}}}\right)}{\left.{d}{x}\right.}$

Then, express it as difference of two integrals.

$\displaystyle=\int\frac{{1}}{{{3}{x}-{1}}}{\left.{d}{x}\right.}-\int\frac{{1}}{{{3}{x}+{1}}}{\left.{d}{x}\right.}$

To evaluate each integral, apply substitution method.

$\displaystyle{u}={3}{x}-{1}$
$\displaystyle{d}{u}={3}{\left.{d}{x}\right.}$
$\displaystyle\frac{{1}}{{3}}{d}{u}={\left.{d}{x}\right.}$
$\displaystyle{w}={3}{x}+{1}$
$\displaystyle{d}{w}={3}{\left.{d}{x}\right.}$
$\displaystyle\frac{{1}}{{3}}{d}{w}={\left.{d}{x}\right.}$

Expressing the two integrals in terms of u and w, it becomes:

$\displaystyle=\int\frac{{1}}{{u}}\cdot\frac{{1}}{{3}}{d}{u}-\int\frac{{1}}{{w}}\cdot\frac{{1}}{{3}}{d}{w}$

$\displaystyle=\frac{{1}}{{3}}\int\frac{{1}}{{u}}{d}{u}-\frac{{1}}{{3}}\int\frac{{1}}{{w}}{d}{w}$

To take the integral of this, apply the formula $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle=\frac{{1}}{{3}}{\ln}{\left|{u}\right|}-\frac{{1}}{{3}}{\ln}{\left|{w}\right|}+{C}$

And, substitute back u=3x-1 and w=3x+1.

$\displaystyle=\frac{{1}}{{3}}{\ln}{\left|{3}{x}-{1}\right|}-\frac{{1}}{{3}}{\ln}{\left|{3}{x}+{1}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{2}}{{{9}{{x}}^{{2}}-{1}}}=\frac{{1}}{{3}}{\ln}{\left|{3}{x}-{1}\right|}-\frac{{1}}{{3}}{\ln}{\left|{3}{x}+{1}\right|}+{C}$ .