Notes: `int 5/(x^2+3x-4) dx` Use partial fractions to find the indefinite integral

Thursday, 22 June 2017

$\displaystyle\int\frac{{5}}{{{{x}}^{{2}}+{3}{x}-{4}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

$\displaystyle\int\frac{{5}}{{{{x}}^{{2}}+{3}{x}-{4}}}{\left.{d}{x}\right.}$

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

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

Then, express it as sum of fractions.

$\displaystyle\frac{{5}}{{{\left({x}+{4}\right)}{\left({x}-{1}\right)}}}=\frac{{A}}{{{x}+{4}}}+\frac{{B}}{{{x}-{1}}}$

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

$\displaystyle{\left({x}+{4}\right)}{\left({x}-{1}\right)}\cdot\frac{{5}}{{{\left({x}+{4}\right)}{\left({x}-{1}\right)}}}={\left(\frac{{A}}{{{x}+{4}}}+\frac{{B}}{{{x}-{1}}}\right)}\cdot{\left({x}+{4}\right)}{\left({x}-{1}\right)}$

$\displaystyle{5}={A}{\left({x}-{1}\right)}+{B}{\left({x}+{4}\right)}$

Then, assign values to x in which either x+4 or x-1 will become zero.

So plug-in x=-4 to get the value of A.

$\displaystyle{5}={A}{\left(-{4}-{1}\right)}+{B}{\left(-{4}+{4}\right)}$

$\displaystyle{5}={A}{\left(-{5}\right)}+{B}{\left({0}\right)}$

$\displaystyle{5}=-{5}{A}$

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

Also, plug-in x=1

$\displaystyle{5}={A}{\left({1}-{1}\right)}+{B}{\left({1}+{4}\right)}$

...

$\displaystyle\int\frac{{5}}{{{{x}}^{{2}}+{3}{x}-{4}}}{\left.{d}{x}\right.}$

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

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

Then, express it as sum of fractions.

$\displaystyle\frac{{5}}{{{\left({x}+{4}\right)}{\left({x}-{1}\right)}}}=\frac{{A}}{{{x}+{4}}}+\frac{{B}}{{{x}-{1}}}$

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

$\displaystyle{5}={A}{\left({x}-{1}\right)}+{B}{\left({x}+{4}\right)}$

Then, assign values to x in which either x+4 or x-1 will become zero.

So plug-in x=-4 to get the value of A.

$\displaystyle{5}={A}{\left(-{4}-{1}\right)}+{B}{\left(-{4}+{4}\right)}$

$\displaystyle{5}={A}{\left(-{5}\right)}+{B}{\left({0}\right)}$

$\displaystyle{5}=-{5}{A}$

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

Also, plug-in x=1

$\displaystyle{5}={A}{\left({1}-{1}\right)}+{B}{\left({1}+{4}\right)}$

$\displaystyle{5}={A}{\left({0}\right)}+{B}{\left({5}\right)}$

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

$\displaystyle{1}={B}$

So the partial fraction decomposition of the integrand is

$\displaystyle\int\frac{{5}}{{{{x}}^{{2}}+{3}{x}-{4}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{5}}{{{\left({x}+{4}\right)}{\left({x}-{1}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left(-\frac{{1}}{{{x}+{4}}}+\frac{{1}}{{{x}-{1}}}\right)}{\left.{d}{x}\right.}$

Then, express it as two integrals.

$\displaystyle=\int-\frac{{1}}{{{x}+{4}}}{\left.{d}{x}\right.}+\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}$

$\displaystyle=-\int\frac{{1}}{{{x}+{4}}}+\int\frac{{1}}{{{x}-{1}}}{\left.{d}{x}\right.}$

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

$\displaystyle=-{\ln}{\left|{x}+{4}\right|}+{\ln}{\left|{x}-{1}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{5}}{{{{x}}^{{2}}+{3}{x}-{4}}}{\left.{d}{x}\right.}=-{\ln}{\left|{x}+{4}\right|}+{\ln}{\left|{x}-{1}\right|}+{C}$ .