Notes: `int x/sqrt(x^2-6x+5)` Complete the square and find the indefinite integral

Tuesday, 5 July 2016

$\displaystyle\int\frac{{x}}{\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}}$ Complete the square and find the indefinite integral

$\displaystyle\int\frac{{x}}{\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}}{\left.{d}{x}\right.}$

Let's complete the square of the denominator of the integrand,

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

Now apply integral substitution: $\displaystyle{u}={x}-{3}$

$\displaystyle\Rightarrow{d}{u}={1}{\left.{d}{x}\right.}$

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

Now apply the sum rule,

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

Now let's evaluate the first integral ,

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

Apply integral substitution: $\displaystyle{v}={{u}}^{{2}}-{4}$

$\displaystyle\Rightarrow{d}{v}={2}{u}{d}{u}$

$\displaystyle=\int\frac{{1}}{\sqrt{{{v}}}}\frac{{{d}{v}}}{{2}}$

Take the constant out and apply the power rule,

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

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

$\displaystyle=\sqrt{{{v}}}$

Substitute back $\displaystyle{v}={{u}}^{{2}}-{4}$

$\displaystyle=\sqrt{{{{u}}^{{2}}-{4}}}$

Now let's evaluate the second integral,

$\displaystyle\int\frac{{3}}{\sqrt{{{{u}}^{{2}}-{{2}}^{{2}}}}}{d}{u}$

Take the constant out,

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

Use the standard integral: $\displaystyle\int\frac{{1}}{\sqrt{{{{x}}^{{2}}-{{a}}^{{2}}}}}{\left.{d}{x}\right.}={\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}-{{a}}^{{2}}}}\right|}$

$\displaystyle={3}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{{2}}^{{2}}}}\right|}$ ,

So add the result of the...

$\displaystyle\int\frac{{x}}{\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}}{\left.{d}{x}\right.}$

Let's complete the square of the denominator of the integrand,

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

Now apply integral substitution: $\displaystyle{u}={x}-{3}$

$\displaystyle\Rightarrow{d}{u}={1}{\left.{d}{x}\right.}$

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

Now apply the sum rule,

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

Now let's evaluate the first integral ,

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

Apply integral substitution: $\displaystyle{v}={{u}}^{{2}}-{4}$

$\displaystyle\Rightarrow{d}{v}={2}{u}{d}{u}$

$\displaystyle=\int\frac{{1}}{\sqrt{{{v}}}}\frac{{{d}{v}}}{{2}}$

Take the constant out and apply the power rule,

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

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

$\displaystyle=\sqrt{{{v}}}$

Substitute back $\displaystyle{v}={{u}}^{{2}}-{4}$

$\displaystyle=\sqrt{{{{u}}^{{2}}-{4}}}$

Now let's evaluate the second integral,

$\displaystyle\int\frac{{3}}{\sqrt{{{{u}}^{{2}}-{{2}}^{{2}}}}}{d}{u}$

Take the constant out,

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

Use the standard integral: $\displaystyle\int\frac{{1}}{\sqrt{{{{x}}^{{2}}-{{a}}^{{2}}}}}{\left.{d}{x}\right.}={\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}-{{a}}^{{2}}}}\right|}$

$\displaystyle={3}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{{2}}^{{2}}}}\right|}$ ,

So add the result of the two integrals,

$\displaystyle\sqrt{{{{u}}^{{2}}-{4}}}+{3}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{4}}}\right|}$

Substitute back $\displaystyle{u}={x}-{3}$ and add a constant C to the solution,

$\displaystyle=\sqrt{{{{\left({x}-{3}\right)}}^{{2}}-{4}}}+{3}{\ln}{\left|{x}-{3}+\sqrt{{{{\left({x}-{3}\right)}}^{{2}}-{4}}}\right|}+{C}$

$\displaystyle=\sqrt{{{{x}}^{{2}}-{6}{x}+{9}-{4}}}+{3}{\ln}{\left|{x}-{3}+\sqrt{{{{x}}^{{2}}-{6}{x}+{9}-{4}}}\right|}+{C}$

$\displaystyle=\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}+{3}{\ln}{\left|{x}-{3}+\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}\right|}+{C}$

Notes

Tuesday, 5 July 2016

$\displaystyle\int\frac{{x}}{\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}}$ Complete the square and find the indefinite integral

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Tuesday, 5 July 2016

Complete the square and find the indefinite integral

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle\int\frac{{x}}{\sqrt{{{{x}}^{{2}}-{6}{x}+{5}}}}$ Complete the square and find the indefinite integral

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?