New Notes Blog: `int 1 / (xsqrt(x^4-4)) dx` Find the indefinite integral

Monday, 10 March 2014

$\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

For the given integral: $\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}$ , we may apply u-substitution by letting:

$\displaystyle{u}={{x}}^{{4}}-{4}$ then $\displaystyle{d}{u}={4}{{x}}^{{3}}{\left.{d}{x}\right.}$ .

Rearrange $\displaystyle{d}{u}={4}{{x}}^{{3}}{\left.{d}{x}\right.}$ into $\displaystyle\frac{{{d}{u}}}{{{4}{{x}}^{{3}}}}={\left.{d}{x}\right.}$

Plug-in $\displaystyle{u}={{x}}^{{4}}-{4}$ and $\displaystyle\frac{{{d}{u}}}{{{4}{{x}}^{{3}}}}={\left.{d}{x}\right.}$ , we get:

$\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{x}\sqrt{{{u}}}}}\cdot\frac{{{d}{u}}}{{{4}{{x}}^{{3}}}}$

$\displaystyle=\int\frac{{1}}{{{4}{{x}}^{{4}}\sqrt{{{u}}}}}{d}{u}$

Recall $\displaystyle{u}={{x}}^{{4}}-{4}$ then adding 4 on both sides becomes: $\displaystyle{u}+{4}={{x}}^{{4}}$ .

Plug-in $\displaystyle{{x}}^{{4}}\ldots$

For the given integral: $\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}$ , we may apply u-substitution by letting:

$\displaystyle{u}={{x}}^{{4}}-{4}$ then $\displaystyle{d}{u}={4}{{x}}^{{3}}{\left.{d}{x}\right.}$ .

Rearrange $\displaystyle{d}{u}={4}{{x}}^{{3}}{\left.{d}{x}\right.}$ into $\displaystyle\frac{{{d}{u}}}{{{4}{{x}}^{{3}}}}={\left.{d}{x}\right.}$

Plug-in $\displaystyle{u}={{x}}^{{4}}-{4}$ and $\displaystyle\frac{{{d}{u}}}{{{4}{{x}}^{{3}}}}={\left.{d}{x}\right.}$ , we get:

$\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{x}\sqrt{{{u}}}}}\cdot\frac{{{d}{u}}}{{{4}{{x}}^{{3}}}}$

$\displaystyle=\int\frac{{1}}{{{4}{{x}}^{{4}}\sqrt{{{u}}}}}{d}{u}$

Recall $\displaystyle{u}={{x}}^{{4}}-{4}$ then adding 4 on both sides becomes: $\displaystyle{u}+{4}={{x}}^{{4}}$ .

Plug-in $\displaystyle{{x}}^{{4}}={u}+{4}$ in the integral:

$\displaystyle\int\frac{{1}}{{{4}{{x}}^{{4}}\sqrt{{{u}}}}}{d}{u}$ = $\displaystyle\int\frac{{1}}{{{4}{\left({u}+{4}\right)}\sqrt{{{u}}}}}{d}{u}$

Apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ :

$\displaystyle\int\frac{{1}}{{{4}{\left({u}+{4}\right)}\sqrt{{{u}}}}}{d}{u}=\frac{{1}}{{4}}\int\frac{{1}}{{{\left({u}+{4}\right)}\sqrt{{{u}}}}}{d}{u}$

Apply another set of substitution by letting:

$\displaystyle{v}=\sqrt{{{u}}}$ which is the same as $\displaystyle{{v}}^{{2}}={u}$ .

Then taking the derivative on both sides, we get $\displaystyle{2}{v}{d}{v}={d}{u}$ .

Plug-in $\displaystyle{u}={{v}}^{{2}}$ , $\displaystyle{d}{u}={2}{v}{d}{v}$ , and $\displaystyle\sqrt{{{u}}}={v}$ , we get:

$\displaystyle\frac{{1}}{{4}}\int\frac{{1}}{{{\left({u}+{4}\right)}\sqrt{{{u}}}}}{d}{u}=\frac{{1}}{{4}}\int\frac{{1}}{{{\left({{v}}^{{2}}+{4}\right)}{v}}}{\left({2}{v}{d}{v}\right)}$

We simplify by cancelling out common factors v and 2:

$\displaystyle\frac{{1}}{{4}}\int\frac{{1}}{{{\left({{v}}^{{2}}+{4}\right)}{v}}}{\left({2}{v}{d}{v}\right)}=\frac{{1}}{{2}}\int\frac{{{d}{v}}}{{{{v}}^{{2}}+{4}}}{\quad\text{or}\quad}\frac{{1}}{{2}}\int\frac{{{d}{v}}}{{{{v}}^{{2}}+{{2}}^{{2}}}}$

The integral part resembles the integration formula:

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

Then,

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

$\displaystyle=\frac{{1}}{{4}}{\arctan{{\left(\frac{{v}}{{2}}\right)}}}+{C}$

Recall that we let $\displaystyle{v}=\sqrt{{{u}}}$ and $\displaystyle{u}={{x}}^{{4}}-{4}$ then $\displaystyle{v}=\sqrt{{{{x}}^{{4}}-{4}}}$

Plug-in $\displaystyle{v}=\sqrt{{{{x}}^{{4}}-{4}}}$ in $\displaystyle\frac{{1}}{{4}}{\arctan{{\left(\frac{{v}}{{2}}\right)}}}+{C}$ to get the final answer:

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

New Notes Blog

Monday, 10 March 2014

$\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}$ 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, 10 March 2014

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?

$\displaystyle\int\frac{{1}}{{{x}\sqrt{{{{x}}^{{4}}-{4}}}}}{\left.{d}{x}\right.}$ 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?