`int 2x/(x^2+6x+13) dx` Find or evaluate the integral by completing the square

Monday, June 15, 2009

$\displaystyle\int{2}\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

For the given integral: $\displaystyle\int{2}\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}$ , we may apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int{2}\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}={2}\int\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}$

To be able to evaluate this, we apply completing the square on $\displaystyle{{x}}^{{2}}+{6}{x}+{13}$ .

The $\displaystyle{{x}}^{{2}}+{6}{x}+{13}$ resembles $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ where:

$\displaystyle{a}={1}$ and $\displaystyle{b}={6}$ that we can plug-into $\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}$ .

$\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}={{\left(-\frac{{{6}}}{{{2}\cdot{1}}}\right)}}^{{2}}$

$\displaystyle={{\left(-\frac{{6}}{{2}}\right)}}^{{2}}$

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

$\displaystyle={9}$

To complete the square, we add and subtract 9:

$\displaystyle{{x}}^{{2}}+{6}{x}+{13}+{9}-{9}$

Group them as: $\displaystyle{\left({{x}}^{{2}}+{6}{x}+{9}\right)}-{9}+{13}$

Simplify: $\displaystyle{\left({{x}}^{{2}}+{6}{x}+{9}\right)}+{4}$

Apply factoring for the perfect square trinomial: $\displaystyle{{x}}^{{2}}+{6}{x}+{9}={{\left({x}+{3}\right)}}^{{2}}$

$\displaystyle{\left({{x}}^{{2}}+{6}{x}+{9}\right)}+{4}={{\left({x}+{3}\right)}}^{{2}}+{4}$

Which means $\displaystyle{{x}}^{{2}}+{6}{x}+{13}={{\left({x}+{3}\right)}}^{{2}}+{4}$ then the integral becomes:

$\displaystyle{2}\int\frac{{x}}{\sqrt{{{{x}}^{{2}}+{6}{x}+{13}}}}{\left.{d}{x}\right.}={2}\int\frac{{x}}{{{{\left({x}+{3}\right)}}^{{2}}+{4}}}{\left.{d}{x}\right.}$

For the integral part, we apply u-substitution by letting:

$\displaystyle{u}={x}+{3}$ then $\displaystyle{x}={u}-{3}$ and $\displaystyle{d}{u}={\left.{d}{x}\right.}$

Then,

$\displaystyle{2}\int\frac{{x}}{{{{\left({x}+{3}\right)}}^{{2}}+{4}}}{\left.{d}{x}\right.}={2}\int\frac{{{u}-{3}}}{{{{u}}^{{2}}+{4}}}{d}{u}$

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

$\displaystyle{2}\int\frac{{{u}-{3}}}{{{{u}}^{{2}}+{4}}}{d}{u}={2}{\left[\int\frac{{u}}{{{{u}}^{{2}}+{4}}}{d}{u}-\int\frac{{3}}{{{{u}}^{{2}}+{4}}}{d}{u}\right]}$

For the integration of $\displaystyle\int\frac{{u}}{{{{u}}^{{2}}+{4}}}{d}{u}$ , let:

$\displaystyle{v}={{u}}^{{2}}+{4}$ then $\displaystyle{d}{v}={2}{u}{d}{u}$ or $\displaystyle\frac{{{d}{v}}}{{2}}={u}{d}{u}$ .

Then,

$\displaystyle\int\frac{{u}}{{{{u}}^{{2}}+{4}}}{d}{u}=\int\frac{{\frac{{{d}{v}}}{{2}}}}{{{v}}}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{{d}{v}}}{{{v}}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{v}\right|}+{C}$

Plug-in $\displaystyle{v}={{u}}^{{2}}+{4},$ we get: $\displaystyle\int\frac{{u}}{{{{u}}^{{2}}+{4}}}{d}{u}=\frac{{1}}{{2}}{\ln}{\left|{{u}}^{{2}}+{4}\right|}+{C}$

For the second integration: $\displaystyle-\int\frac{{3}}{{{{u}}^{{2}}+{4}}}{d}{u}$ , we follow the basic integration formula for inverse tangent function:

$\displaystyle\int\frac{{{d}{u}}}{{{{u}}^{{2}}+{{a}}^{{2}}}}=\frac{{1}}{{a}}{\arctan{{\left(\frac{{u}}{{a}}\right)}}}+{C}$

Then,

$\displaystyle-\int\frac{{3}}{{{{u}}^{{2}}+{4}}}{d}{u}=-{3}\int\frac{{{d}{u}}}{{{{u}}^{{2}}+{{2}}^{{2}}}}$

$\displaystyle=-{3}\cdot\frac{{1}}{{2}}{\arctan{{\left(\frac{{u}}{{2}}\right)}}}+{C}$

$\displaystyle=-\frac{{3}}{{2}}{\arctan{{\left(\frac{{u}}{{2}}\right)}}}+{C}$

Combine the results, we get:

$\displaystyle{2}{\left[\int{\left(\frac{{u}}{{{{u}}^{{2}}+{4}}}{d}{u}-\int\frac{{3}}{{{{u}}^{{2}}+{4}}}{d}{u}\right]}\right.}$

$\displaystyle={2}\cdot{\left[\frac{{1}}{{2}}{\ln}{\left|{{u}}^{{2}}+{4}\right|}-\frac{{3}}{{2}}{\arctan{{\left(\frac{{u}}{{2}}\right)}}}\right]}+{C}$

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

Plug-in $\displaystyle{u}={x}+{3}$ to solve for the final answer:

$\displaystyle\int{2}\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}={\ln}{\left|{{\left({x}+{3}\right)}}^{{2}}+{4}\right|}-{3}{\arctan{{\left(\frac{{{x}+{3}}}{{2}}\right)}}}+{C}$

Blog

Monday, June 15, 2009

$\displaystyle\int{2}\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

No comments:

Post a Comment

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Monday, June 15, 2009

Find or evaluate the integral by completing the square

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle\int{2}\frac{{x}}{{{{x}}^{{2}}+{6}{x}+{13}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?