New Notes Blog: `int 1/(xsqrt(9x^2+1)) dx` Find the indefinite integral

Monday, 25 November 2013

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

Recall that indefinite integral follows the formula: $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

For the given problem $\displaystyle\int\frac{{1}}{{{x}\sqrt{{{9}{{x}}^{{2}}+{1}}}}}{\left.{d}{x}\right.}$ , it resembles one of the formula from integration table. We may apply the integral formula for rational function with roots as:

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

Recall that indefinite integral follows the formula: $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

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

For easier comparison, we apply u-substitution by letting: $\displaystyle{{u}}^{{2}}={9}{{x}}^{{2}}$ or $\displaystyle{{\left({3}{x}\right)}}^{{2}}$ then $\displaystyle{u}={3}{x}$ or $\displaystyle\frac{{u}}{{3}}={x}$ .

Note: The corresponding value of $\displaystyle{{a}}^{{2}}={1}$ or $\displaystyle{{1}}^{{2}}$ then $\displaystyle{a}={1}$ .

For the derivative of $\displaystyle{u}$ , we get: $\displaystyle{d}{u}={3}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{3}}={\left.{d}{x}\right.}$ .

Plug-in the values on the integral problem, we get:

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

$\displaystyle=\int\frac{{3}}{{{u}\sqrt{{{{u}}^{{2}}+{1}}}}}\cdot\frac{{{d}{u}}}{{3}}$

$\displaystyle=\int\frac{{{d}{u}}}{{{u}\sqrt{{{{u}}^{{2}}+{1}}}}}$

Applying the aforementioned integral formula where $\displaystyle{{a}}^{{2}}={1}$ and $\displaystyle{a}={1}$ , we get:

$\displaystyle\int\frac{{{d}{u}}}{{{u}\sqrt{{{{u}}^{{2}}+{1}}}}}=-\frac{{1}}{{1}}{\ln{{\left(\frac{{{1}+\sqrt{{{{u}}^{{2}}+{1}}}}}{{u}}\right)}}}+{C}$

$\displaystyle=-{\ln{{\left(\frac{{{1}+\sqrt{{{{u}}^{{2}}+{1}}}}}{{u}}\right)}}}+{C}$

$\displaystyle={\ln{{\left({{\left(\frac{{{1}+\sqrt{{{{u}}^{{2}}+{1}}}}}{{u}}\right)}}^{{-{{1}}}}\right)}}}+{C}$

$\displaystyle={\ln{{\left(\frac{{u}}{{{1}+\sqrt{{{{u}}^{{2}}+{1}}}}}\right)}}}+{C}$

Plug-in $\displaystyle{{u}}^{{2}}={9}{{x}}^{{2}}$ and $\displaystyle{u}={3}{x}$ and we get the indefinite integral as:

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

New Notes Blog

Monday, 25 November 2013

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

Monday, 25 November 2013

Find the indefinite integral

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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?

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