`int (x^2+3)/(xsqrt(x^2-4)) dx` Find the indefinite integral

Friday, February 24, 2012

$\displaystyle\int\frac{{{{x}}^{{2}}+{3}}}{{{x}\sqrt{{{{x}}^{{2}}-{4}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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

where:

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

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as the constant of integration.

The given problem: $\displaystyle\int\frac{{{{x}}^{{2}}+{3}}}{{{x}\sqrt{{{{x}}^{{2}}-{4}}}}}{\left.{d}{x}\right.}$ has an integrand of $\displaystyle{f{{\left({x}\right)}}}=\frac{{{{x}}^{{2}}+{3}}}{{{x}\sqrt{{{{x}}^{{2}}-{4}}}}}$ .

Apply u-substitution on $\displaystyle{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ by letting $\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$ or $\displaystyle{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{{2}{x}}}$ :

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

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

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

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

$\displaystyle\int\frac{{{\left({u}+{3}\right)}{d}{u}}}{{{2}{u}\sqrt{{{u}-{4}}}}}={\left(\frac{{1}}{{2}}\right)}\int\frac{{{\left({u}+{3}\right)}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}$

Apply the basic integration property for sum:

$\displaystyle\int{\left({u}+{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}+\int{\left({v}\right)}{\left.{d}{x}\right.}.$

$\displaystyle{\left(\frac{{1}}{{2}}\right)}\int\frac{{{\left({u}+{3}\right)}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}={\left(\frac{{1}}{{2}}\right)}{\left[\int\frac{{{u}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}+\int\frac{{{3}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}\right]}$

For the integration of the $\displaystyle\int\frac{{{u}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}$ , we can cancel out the u:

$\displaystyle\int\frac{{{u}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}=\int\frac{{{d}{u}}}{\sqrt{{{u}-{4}}}}$

Let $\displaystyle{v}={u}-{4}$ then $\displaystyle{d}{v}={d}{u}$ .

Apply the Law of exponents: $\displaystyle\sqrt{{{x}}}=\frac{{{x}}^{{1}}}{{2}}$ and $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{{n}}}}$ , we get:

$\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{u}-{4}}}}=\int\frac{{{d}{v}}}{\sqrt{{{v}}}}$

Apply the Power Rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$

$\displaystyle\int{{v}}^{{-\frac{{1}}{{2}}}}{d}{v}=\frac{{{v}}^{{\frac{{-{1}}}{{2}}+{1}}}}{{\frac{{-{1}}}{{2}}+{1}}}+{C}$

$\displaystyle=\frac{{{v}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}$

$\displaystyle={{v}}^{{\frac{{1}}{{2}}}}\cdot{\left(\frac{{2}}{{1}}\right)}$

$\displaystyle={2}{{v}}^{{\frac{{1}}{{2}}}}$ or $\displaystyle{2}\sqrt{{{v}}}$

With $\displaystyle{v}={u}-{4}$ then $\displaystyle{2}\sqrt{{{v}}}={2}\sqrt{{{u}-{4}}}$ .

The integral becomes:

$\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{u}-{4}}}}={2}\sqrt{{{u}-{4}}}.$

For the integration of $\displaystyle\int\frac{{{3}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}$ , we basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}$

$\displaystyle\int\frac{{{3}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}={3}\int\frac{{{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}$

Let: $\displaystyle{v}=\sqrt{{{u}-{4}}}$

Then square both sides to get $\displaystyle{{v}}^{{2}}={u}-{4}$ then $\displaystyle{{v}}^{{2}}+{4}={u}$

Applying implicit differentiation on $\displaystyle{{v}}^{{2}}={u}-{4}$ , we get: $\displaystyle{2}{v}{d}{v}={d}{u}$ .

Plug-in $\displaystyle{d}{u}={2}{v}{d}{v}$ , $\displaystyle{u}={{v}}^{{2}}+{4}$ and $\displaystyle{v}=\sqrt{{{u}-{4}}}$ , we get:

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

$\displaystyle={3}\int\frac{{{2}{d}{v}}}{{{\left({{v}}^{{2}}+{4}\right)}}}$

$\displaystyle={3}\cdot{2}\int\frac{{{d}{v}}}{{{{v}}^{{2}}+{4}}}$

$\displaystyle={6}\int\frac{{{d}{v}}}{{{{v}}^{{2}}+{4}}}$

The integral part resembles the basic integration for inverse tangent function:

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

Then,

$\displaystyle{6}\int\frac{{{d}{v}}}{{{{v}}^{{2}}+{4}}}={6}\cdot{\left(\frac{{1}}{{2}}\right)}{\arctan{{\left(\frac{{v}}{{2}}\right)}}}+{C}$

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

Plug-in $\displaystyle{v}=\sqrt{{{u}-{4}}}$ , we get:

$\displaystyle\int\frac{{{3}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}={3}{\arctan{{\left(\frac{\sqrt{{{u}-{4}}}}{{2}}\right)}}}+{C}$

Combining the results, we get:

$\displaystyle{\left(\frac{{1}}{{2}}\right)}{\left[\int\frac{{{u}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}+\int\frac{{{3}{d}{u}}}{{{u}\sqrt{{{u}-{4}}}}}\right]}={\left(\frac{{1}}{{2}}\right)}\cdot{\left[{2}\sqrt{{{u}-{4}}}+{3}{\arctan{{\left(\frac{\sqrt{{{u}-{4}}}}{{2}}\right)}}}\right]}+{C}$

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

Plug-in $\displaystyle{u}={{x}}^{{2}}$ to get the final answer:

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

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Friday, February 24, 2012

$\displaystyle\int\frac{{{{x}}^{{2}}+{3}}}{{{x}\sqrt{{{{x}}^{{2}}-{4}}}}}{\left.{d}{x}\right.}$ 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?

Friday, February 24, 2012

Find the indefinite integral

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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?

$\displaystyle\int\frac{{{{x}}^{{2}}+{3}}}{{{x}\sqrt{{{{x}}^{{2}}-{4}}}}}{\left.{d}{x}\right.}$ 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?