`int 12/(1+9x^2) dx` Find the indefinite integral

Monday, April 22, 2013

$\displaystyle\int\frac{{12}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Recall that the indefinite integral is denoted as:

$\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

There properties and basic formulas of integration we can apply to simply certain function.

For the problem $\displaystyle\int\frac{{{12}}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

we apply the $\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ to become:

$\displaystyle{12}\int\frac{{1}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

Then apply the basic inverse trigonometric function formula:

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

By comparison with the basic formula and the given problem, we can let:

$\displaystyle{{a}}^{{2}}={1}$

$\displaystyle{{u}}^{{2}}={9}{{x}}^{{2}}{\quad\text{or}\quad}{{\left({3}{x}\right)}}^{{2}}$

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

To satisfy the given formula, we need to multiply the integral by $\displaystyle\frac{{3}}{{3}}$ to

be able to match $\displaystyle{d}{u}={3}{\left.{d}{x}\right.}$ .

The integral value will note change since multiplying by 3/3 is the same as multiplying by 1. Note: $\displaystyle\frac{{3}}{{3}}={1}$ and $\displaystyle\frac{{3}}{{3}}={3}\cdot{\left(\frac{{1}}{{3}}\right)}$

Then $\displaystyle{12}\int\frac{{1}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}\cdot\frac{{3}}{{3}}$

$\displaystyle={12}\int\frac{{1}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}\cdot{3}\cdot\frac{{1}}{{3}}$

$\displaystyle={12}{\left(\frac{{1}}{{3}}\right)}\int{1}\cdot\frac{{3}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle={4}\int\frac{{{3}{\left.{d}{x}\right.}}}{{{1}+{9}{{x}}^{{2}}}}$

The $\displaystyle\int\frac{{{3}{\left.{d}{x}\right.}}}{{{1}+{9}{{x}}^{{2}}}}$ is now similar to $\displaystyle\int\frac{{{d}{u}}}{{{{a}}^{{2}}+{{u}}^{{2}}}}$ where:

$\displaystyle{d}{u}={3}{\left.{d}{x}\right.}$ , $\displaystyle{{a}}^{{2}}={1}$ and $\displaystyle{{u}}^{{2}}={9}{{x}}^{{2}}{\quad\text{or}\quad}{{\left({3}{x}\right)}}^{{2}}$

then $\displaystyle{a}={1}$ and $\displaystyle{u}={3}{x}$ .

Plug-in $\displaystyle{a}={1}$ and $\displaystyle{u}={3}{x}$ in $\displaystyle\frac{{1}}{{a}}{\arctan{{\left(\frac{{u}}{{a}}\right)}}}+{C}$ , we get:

$\displaystyle{4}\cdot\int\frac{{{3}{\left.{d}{x}\right.}}}{{{1}+{9}{{x}}^{{2}}}}={4}\cdot\frac{{1}}{{1}}{\arctan{{\left(\frac{{{3}{x}}}{{1}}\right)}}}+{C}$

$\displaystyle={4}{\arctan{{\left({3}{x}\right)}}}+{C}$

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Monday, April 22, 2013

$\displaystyle\int\frac{{12}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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

Monday, April 22, 2013

Find the indefinite integral

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

$\displaystyle\int\frac{{12}}{{{1}+{9}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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