New Notes Blog: `int (6x)/(x^3-8) dx` Use partial fractions to find the indefinite integral

For the given integral problem: $\displaystyle\int\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}{\left.{d}{x}\right.}$ , we may partial fraction decomposition to expand the integrand: $\displaystyle{f{{\left({x}\right)}}}=\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}$ .

The pattern on setting up partial fractions will depend on the factors of the denominator. For the given problem, the denominator is in a form of difference of perfect cube : $\displaystyle{{x}}^{{3}}-{{y}}^{{3}}={\left({x}-{y}\right)}{\left({{x}}^{{2}}+{x}{y}+{{y}}^{{2}}\right)}$

Applying the special factoring on $\displaystyle{\left({{x}}^{{3}}-{8}\right)}$ , we get:

$\displaystyle{\left({{x}}^{{3}}-{8}\right)}={\left({{x}}^{{3}}-{{2}}^{{3}}\right)}$

$\displaystyle={\left({x}-{2}\right)}{\left({{x}}^{{2}}+{x}\cdot{2}+{{2}}^{{2}}\right)}$

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

For the linear factor $\displaystyle{\left({x}-{2}\right)}$ , we will have partial fraction: $\displaystyle\frac{{A}}{{{x}-{2}}}$ .

For the quadratic factor $\displaystyle{\left({{x}}^{{2}}+{2}{x}+{4}\right)}$ , we will have partial fraction: $\displaystyle\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}+{2}{x}+{4}}}$ .

The integrand becomes:

$\displaystyle\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}=\frac{{A}}{{{x}-{2}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}+{2}{x}+{4}}}$

Multiply both side by the $\displaystyle{L}{C}{D}={\left({x}-{2}\right)}{\left({{x}}^{{2}}+{2}{x}+{4}\right)}$ :

$\displaystyle{\left(\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}\right)}\cdot{\left({x}-{2}\right)}{\left({{x}}^{{2}}+{2}{x}+{4}\right)}={\left[\frac{{A}}{{{x}-{2}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right]}\cdot{\left({x}-{2}\right)}{\left({{x}}^{{2}}+{2}{x}+{4}\right)}$

$\displaystyle{6}{x}={A}{\left({{x}}^{{2}}+{2}{x}+{4}\right)}+{\left({B}{x}+{C}\right)}{\left({x}-{2}\right)}$

We apply zero-factor property on $\displaystyle{\left({x}-{2}\right)}{\left({{x}}^{{2}}+{2}{x}+{4}\right)}$ to solve for values we can assign on x.

$\displaystyle{x}-{2}={0}$ then $\displaystyle{x}={2}$

$\displaystyle{{x}}^{{2}}+{2}{x}+{4}={0}$ then $\displaystyle{x}=-{1}\pm\sqrt{{{3}}}{i}$

To solve for $\displaystyle{A}$ , we plug-in $\displaystyle{x}={2}$ :

$\displaystyle{6}\cdot{2}={A}{\left({{2}}^{{2}}+{2}\cdot{2}+{4}\right)}+{\left({B}\cdot{2}+{C}\right)}{\left({2}-{2}\right)}$

$\displaystyle{12}={A}{\left({4}+{4}+{4}\right)}+{\left({2}{B}+{C}\right)}{\left({0}\right)}$

$\displaystyle{12}={12}{A}+{0}$

$\displaystyle\frac{{12}}{{12}}=\frac{{{12}{A}}}{{12}}$

$\displaystyle{A}={1}$

To solve for $\displaystyle{C}$ , plug-in $\displaystyle{A}={1}$ and $\displaystyle{x}={0}$ so that $\displaystyle{B}\cdot{x}$ becomes $\displaystyle{0}$ :

$\displaystyle{6}\cdot{0}={A}{\left({{0}}^{{2}}+{2}\cdot{0}+{4}\right)}+{\left({B}\cdot{0}+{C}\right)}{\left({0}-{2}\right)}$

$\displaystyle{0}={1}{\left({0}+{0}+{4}\right)}+{\left({0}+{C}\right)}{\left(-{2}\right)}$

$\displaystyle{0}={4}-{2}{C}$

$\displaystyle{2}{C}={4}$

$\displaystyle\frac{{{2}{C}}}{{2}}=\frac{{4}}{{2}}$

$\displaystyle{C}={2}$

To solve for $\displaystyle{B}$ , plug-in $\displaystyle{A}={1}$ , $\displaystyle{C}={2}$ , and $\displaystyle{x}={1}$ :

$\displaystyle{6}\cdot{1}={1}{\left({{1}}^{{2}}+{2}\cdot{1}+{4}\right)}+{\left({B}\cdot{1}+{2}\right)}{\left({1}-{2}\right)}$

$\displaystyle{6}={1}+{2}+{4}+{\left({B}+{2}\right)}\cdot{\left(-{1}\right)}$

$\displaystyle{6}={1}+{2}+{4}-{B}-{2}$

$\displaystyle{6}={5}-{B}$

$\displaystyle{6}-{5}=-{B}$

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

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

Plug-in $\displaystyle{A}={1}$ , $\displaystyle{B}=-{1},$ and $\displaystyle{C}={2}$ , we get the partial fraction decomposition:

$\displaystyle\int\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}{\left.{d}{x}\right.}=\int{\left[\frac{{1}}{{{x}-{2}}}+\frac{{-{x}+{2}}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[\frac{{1}}{{{x}-{2}}}-\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{4}}}+\frac{{2}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right]}{\left.{d}{x}\right.}$

Apply the basic integration property: $\displaystyle\int{\left({u}\pm{v}\pm{w}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}\pm\int{\left({v}\right)}{\left.{d}{x}\right.}\pm\int{\left({w}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle\int{\left[\frac{{1}}{{{x}-{2}}}-\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{4}}}+\frac{{2}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right]}{\left.{d}{x}\right.}=\int\frac{{1}}{{{x}-{2}}}{\left.{d}{x}\right.}-\int\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{4}}}{\left.{d}{x}\right.}+\int\frac{{2}}{{{{x}}^{{2}}+{2}{x}+{4}}}{\left.{d}{x}\right.}$

For the first integral, we apply integration formula for logarithm: $\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}+{C}$ .

Let $\displaystyle{u}={x}-{2}$ then $\displaystyle{d}{u}={\left.{d}{x}\right.}$

$\displaystyle\int\frac{{1}}{{{x}-{2}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{u}}{d}{u}$

$\displaystyle={\ln}{\left|{u}\right|}$

$\displaystyle={\ln}{\left|{x}-{2}\right|}$

For the second integral, we apply indefinite integration formula for rational function:

$\displaystyle\int\frac{{x}}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}{\left.{d}{x}\right.}=\frac{{1}}{{{2}{a}}}{\ln}{\left|{a}{{x}}^{{2}}+{b}{x}+{c}\right|}-\frac{{b}}{{{a}\sqrt{{{4}{a}{c}-{{b}}^{{2}}}}}}{\arctan{{\left(\frac{{{2}{a}{x}+{b}}}{\sqrt{{{4}{a}{c}-{{b}}^{{2}}}}}\right)}}}$

By comparing " $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ " with " $\displaystyle{{x}}^{{2}}+{2}{x}+{4}$ ", we determine the corresponding values: $\displaystyle{a}={1}$ , $\displaystyle{b}={2}$ , and $\displaystyle{c}={4}$ .

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{2}{x}+{4}}}{\left.{d}{x}\right.}=\frac{{1}}{{{2}\cdot{1}}}{\ln}{\left|{1}{{x}}^{{2}}+{2}{x}+{4}\right|}-\frac{{2}}{{{1}\sqrt{{{4}\cdot{1}\cdot{4}-{{2}}^{{2}}}}}}{\arctan{{\left(\frac{{{2}\cdot{1}{x}+{2}}}{\sqrt{{{4}\cdot{1}\cdot{4}-{{2}}^{{2}}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}-\frac{{2}}{\sqrt{{{16}-{4}}}}{\arctan{{\left(\frac{{{2}{x}+{2}}}{\sqrt{{{16}-{4}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}-\frac{{2}}{\sqrt{{{12}}}}{\arctan{{\left(\frac{{{2}{x}+{2}}}{\sqrt{{{12}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}-\frac{{2}}{{{2}\sqrt{{{3}}}}}{\arctan{{\left(\frac{{{2}{\left({x}+{1}\right)}}}{{{2}\sqrt{{{3}}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}-\frac{{1}}{\sqrt{{{3}}}}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}}}{{2}}-\frac{{{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}$

Apply indefinite integration formula for rational function with $\displaystyle{a}={1}$ , $\displaystyle{b}={2}$ , and $\displaystyle{c}={4}$ :

$\displaystyle\int\frac{{1}}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}{\left.{d}{x}\right.}=\frac{{2}}{\sqrt{{{4}{a}{c}-{{b}}^{{2}}}}}{\arctan{{\left(\frac{{{2}{a}{x}+{b}}}{\sqrt{{{4}{a}{c}-{{b}}^{{2}}}}}\right)}}}+{C}$

Then,

$\displaystyle\int\frac{{2}}{{{{x}}^{{2}}+{2}{x}+{4}}}{\left.{d}{x}\right.}={2}\int\frac{{1}}{{{{x}}^{{2}}+{2}{x}+{4}}}{\left.{d}{x}\right.}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{\sqrt{{{4}\cdot{1}\cdot{4}-{{2}}^{{2}}}}}{\arctan{{\left(\frac{{{2}\cdot{1}{x}+{2}}}{\sqrt{{{4}\cdot{1}\cdot{4}-{{2}}^{{2}}}}}\right)}}}\right]}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{\sqrt{{{16}-{4}}}}{\arctan{{\left(\frac{{{2}{x}+{2}}}{\sqrt{{{16}-{4}}}}\right)}}}\right]}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{{{2}\sqrt{{{12}}}}}{\arctan{{\left(\frac{{{2}{x}+{2}}}{\sqrt{{{12}}}}\right)}}}\right]}$

$\displaystyle={2}\cdot{\left[\frac{{2}}{{{2}\sqrt{{{3}}}}}{\arctan{{\left(\frac{{{2}{\left({x}+{1}\right)}}}{{{2}\sqrt{{{3}}}}}\right)}}}\right]}$

$\displaystyle={2}\cdot{\left[\frac{{1}}{\sqrt{{{3}}}}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}\right]}$

$\displaystyle=\frac{{2}}{\sqrt{{{3}}}}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}$

$\displaystyle=\frac{{{2}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}$

Combining the results, we get the indefinite integral as:

$\displaystyle\int\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}{\left.{d}{x}\right.}={\ln}{\left|{x}-{2}\right|}-{\left[\frac{{{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}}}{{2}}-\frac{{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}{\sqrt{{{3}}}}\right]}+\frac{{{2}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}+{C}$

$\displaystyle={\ln}{\left|{x}-{2}\right|}-\frac{{{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}}}{{2}}+\frac{{{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}+\frac{{{2}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}+{C}$

$\displaystyle=\frac{{{2}{\ln}{\left|{x}-{2}\right|}-{\ln}{\left|{{x}}^{{2}}+{2}{x}+{4}\right|}}}{{2}}+\frac{{{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}+{2}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}+{C}$

$\displaystyle=\frac{{{\ln}{\left|\frac{{{\left({x}-{2}\right)}}^{{2}}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right|}}}{{2}}+\frac{{{3}{\arctan{{\left(\frac{{{x}+{1}}}{\sqrt{{{3}}}}\right)}}}}}{\sqrt{{{3}}}}+{C}$

$\displaystyle=\frac{{{\ln}{\left|\frac{{{{x}}^{{2}}-{4}{x}+{4}}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right|}}}{{2}}+\sqrt{{{3}}}{\arctan{{\left(\frac{{\sqrt{{{3}}}{\left({x}+{1}\right)}}}{{3}}\right)}}}+{C}$

$\displaystyle=\frac{{{\ln}{\left|\frac{{{{x}}^{{2}}-{4}{x}+{4}}}{{{{x}}^{{2}}+{2}{x}+{4}}}\right|}}}{{2}}+\sqrt{{{3}}}{\arctan{{\left(\frac{{{x}\sqrt{{{3}}}+\sqrt{{{3}}}}}{{3}}\right)}}}+{C}$

New Notes Blog

Thursday, 23 April 2015

$\displaystyle\int\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Thursday, 23 April 2015

Use partial fractions to find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{{{6}{x}}}{{{{x}}^{{3}}-{8}}}{\left.{d}{x}\right.}$ Use partial fractions to find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?