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Lösung 2.1:5a

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.1:5a moved to Solution 2.1:5a: Robot: moved page)
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(HINT: multiply the top and bottom by the conjugate of the denominator.)
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<center> [[Image:2_1_5a-1(2).gif]] </center>
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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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<center> [[Image:2_1_5a-2(2).gif]] </center>
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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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Thus,
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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,
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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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<math>\begin{align}
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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}\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{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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& =\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.
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This can also be written with square roots as
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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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<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{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{1}{9}\left( x+9 \right)^{\frac{1}{2}}+\frac{1}{9}x^{\frac{1}{2}} \\
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\end{align}</math>

Version vom 13:29, 18. Okt. 2008

(HINT: multiply the top and bottom by the conjugate of the denominator.)

If we multiply top and bottom of the fraction by the conjugate expression, x+9+x , then the conjugate rule gives that denominator's root is squared away:


1x+9−x=1x+9−xx+9+xx+9+x=x+9+xx+92−x2=x+9−xx+9+x=9x+9+x


Thus,


dxx+9−x=91x+9+xdx 


If we write the square roots in power form,


91x+921+x21dx ,

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


91x+921+x21dx=9121+1x+921+1+x21+121+1+C=9123x+923+23x23+C=9132x+923+32x23+C=227x+923+227x23+C,

where C is an arbitrary constant.

This can also be written with square roots as


227x+9x+9+227xx+C 


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


ddx227x+923+227x23+C=22723x+923−1+22723x23−1+0=91x+921+91x21

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