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

Saturday, 26 April 2014

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

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

To solve using the partial fraction method, the first step is to factor the denominator of the integrand.

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

Then, express it as a sum of two fractions.

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

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

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

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

Then, assign values to x in such a way that either (x+3) or (x-3) will be zero. So, plug-in x =...

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

To solve using the partial fraction method, the first step is to factor the denominator of the integrand.

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

Then, express it as a sum of two fractions.

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

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

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

Then, assign values to x in such a way that either (x+3) or (x-3) will be zero. So, plug-in x = 3 to get the value of A.

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

$\displaystyle{1}={A}\cdot{6}+{B}\cdot{0}$

$\displaystyle{1}={6}{A}$

$\displaystyle\frac{{1}}{{6}}={A}$

Also, plug-in x=-3 to get the value of B.

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

$\displaystyle{1}={A}\cdot{0}+{B}\cdot{\left(-{6}\right)}$

$\displaystyle{1}=-{6}{B}$

$\displaystyle-\frac{{1}}{{6}}={B}$

So the partial fraction decomposition of the integrand is:

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{9}}}{\left.{d}{x}\right.}=\int{\left(\frac{{1}}{{{6}{\left({x}-{3}\right)}}}-\frac{{1}}{{{6}{\left({x}+{3}\right)}}}\right)}{\left.{d}{x}\right.}$

Then, express it as difference of two integrals.

$\displaystyle=\int\frac{{1}}{{{6}{\left({x}-{3}\right)}}}{\left.{d}{x}\right.}-\int\frac{{1}}{{{6}{\left({x}+{3}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{6}}\int\frac{{1}}{{{x}-{3}}}{\left.{d}{x}\right.}-\frac{{1}}{{6}}\int\frac{{1}}{{{x}+{3}}}{\left.{d}{x}\right.}$

And, apply the integral formula $\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}+{C}$ .

$\displaystyle=\frac{{1}}{{6}}{\ln}{\left|{x}-{3}\right|}-\frac{{1}}{{6}}{\ln}{\left|{x}+{3}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{9}}}{\left.{d}{x}\right.}=\frac{{1}}{{6}}{\ln}{\left|{x}-{3}\right|}-\frac{{1}}{{6}}{\ln}{\left|{x}+{3}\right|}+{C}$ .