`int sqrt(x^2 - 9)/x^3 dx` Evaluate the integral

Thursday, March 25, 2010

$\displaystyle\int\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}$

Let's evaluate the integral: Apply integration by parts,

$\displaystyle\int{u}{v}'={u}{v}-\int{u}'{v}$

Let $\displaystyle{u}=\sqrt{{{{x}}^{{2}}-{9}}}$

$\displaystyle{v}'=\frac{{1}}{{{x}}^{{3}}}$

$\displaystyle{u}'=\frac{{d}}{{\left.{d}{x}}}{\left(\sqrt{{{{x}}^{{2}}-{9}}}\right)}$

$\displaystyle{u}'=\frac{{1}}{{2}}{{\left({{x}}^{{2}}-{9}\right)}}^{{\frac{{1}}{{2}}-{1}}}{\left({2}{x}\right)}$

$\displaystyle{u}'=\frac{{x}}{\sqrt{{{{x}}^{{2}}-{9}}}}$

$\displaystyle{v}=\int\frac{{1}}{{{x}}^{{3}}}{\left.{d}{x}\right.}$

$\displaystyle{v}=\frac{{{x}}^{{-{3}+{1}}}}{{-{3}+{1}}}$

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

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

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

Apply the integral substitution $\displaystyle{y}=\sqrt{{{{x}}^{{2}}-{9}}}$

$\displaystyle{\left.{d}{y}\right.}=\frac{{1}}{{2}}{{\left({{x}}^{{2}}-{9}\right)}}^{{\frac{{1}}{{2}}-{1}}}{\left({2}{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{\left.{d}{y}\right.}=\frac{{x}}{\sqrt{{{{x}}^{{2}}-{9}}}}{\left.{d}{x}\right.}$

$\displaystyle{\left.{d}{y}\right.}=\frac{{{x}{\left.{d}{x}\right.}}}{{y}}$

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

$\displaystyle=-\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{2}}\int\frac{{\left.{d}{y}}}{{{x}}^{{2}}}$

$\displaystyle=-\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{2}}\int\frac{{\left.{d}{y}}}{{{{y}}^{{2}}+{9}}}$

Now use the standard integral: $\displaystyle\int\frac{{1}}{{{{x}}^{{2}}+{{a}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{a}}{\arctan{{\left(\frac{{x}}{{a}}\right)}}}+{C}$

$\displaystyle=-\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{2}}{\left(\frac{{1}}{{3}}{\arctan{{\left(\frac{{y}}{{3}}\right)}}}\right)}$

Substitute back $\displaystyle{y}=\sqrt{{{{x}}^{{2}}-{9}}}$ and add a constant C to the solution,

$\displaystyle=-\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{6}}{\arctan{{\left(\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{3}}\right)}}}+{C}$

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Thursday, March 25, 2010

$\displaystyle\int\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}$ Evaluate the 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?

Thursday, March 25, 2010

Evaluate the 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?

$\displaystyle\int\frac{\sqrt{{{{x}}^{{2}}-{9}}}}{{{x}}^{{3}}}{\left.{d}{x}\right.}$ Evaluate the integral

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