Lösung 2.1:5a

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(HINT: multiply the top and bottom by the conjugate of the denominator.)
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If we multiply top and bottom of the fraction by the conjugate expression
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<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
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If we multiply top and bottom of the fraction by the conjugate expression,
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<math>\sqrt{x+9}+\sqrt{x}</math>, then the conjugate rule gives that denominator's root is squared away:
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<math>\begin{align}
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& \frac{1}{\sqrt{x+9}-\sqrt{x}}=\frac{1}{\sqrt{x+9}-\sqrt{x}}\centerdot \frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}} \\
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& =\frac{\sqrt{x+9}+\sqrt{x}}{\left( \sqrt{x+9} \right)^{2}-\left( \sqrt{x} \right)^{2}} \\
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& =\frac{\sqrt{x+9}+\sqrt{x}}{x+9-x} \\
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& =\frac{\sqrt{x+9}+\sqrt{x}}{9} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\frac{1}{\sqrt{x+9}-\sqrt{x}}
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&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
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&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt]
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&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt]
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&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.}
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\end{align}</math>}}
Thus,
Thus,
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{{Displayed math||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
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<math>\int{\frac{\,dx}{\sqrt{x+9}-\sqrt{x}}}=\frac{1}{9}\int{\left( \sqrt{x+9}+\sqrt{x} \right)}\,dx</math>
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If we write the square roots in power form,
If we write the square roots in power form,
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{{Displayed math||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
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<math>\frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx</math>,
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we see that we have a standard integral and can write down the primitive functions directly,
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we see that we have a standard integral and can write down the primitive functions directly:
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{{Displayed math||<math>\begin{align}
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\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
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&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]
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<math>\begin{align}
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&= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt]
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& \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx=\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right)+C \\
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&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt]
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& =\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right)+C \\
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&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
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& =\frac{1}{9}\left( \frac{2}{3}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{3}x^{\frac{3}{2}} \right)+C \\
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\end{align}</math>}}
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& =\frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \\
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& \\
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\end{align}</math>,
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where C is an arbitrary constant.
where C is an arbitrary constant.
Zeile 39: Zeile 32:
This can also be written with square roots as
This can also be written with square roots as
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{{Displayed math||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
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<math>\frac{2}{27}\left( x+9 \right)\sqrt{x+9}+\frac{2}{27}x\sqrt{x}+C</math>
 
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To be completely certain that we have everything correctly, we differentiate the answer and see if we get back the integrand:
 
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Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \frac{d}{dx}\left( \frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \right) \\
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\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
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& =\frac{2}{27}\centerdot \frac{3}{2}\left( x+9 \right)^{\frac{3}{2}-1}+\frac{2}{27}\centerdot \frac{3}{2}x^{\frac{3}{2}-1}+0 \\
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&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]
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& =\frac{1}{9}\left( x+9 \right)^{\frac{1}{2}}+\frac{1}{9}x^{\frac{1}{2}} \\
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&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 09:36, 28. Okt. 2008

If we multiply top and bottom of the fraction by the conjugate expression \displaystyle \sqrt{x+9}+\sqrt{x} then the formula for the difference of two squares gives that denominator's root is squared away,

\displaystyle \begin{align}

\frac{1}{\sqrt{x+9}-\sqrt{x}} &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] &= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] &= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] &= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} \end{align}

Thus,

\displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}

If we write the square roots in power form,

\displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,

we see that we have a standard integral and can write down the primitive functions directly,

\displaystyle \begin{align}

\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx &= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt] &= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] &= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] &= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,, \end{align}

where C is an arbitrary constant.

This can also be written with square roots as

\displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}


Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,

\displaystyle \begin{align}

\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr) &= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt] &= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:5a“