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Lösung 2.2:4b

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.2:4b moved to Solution 2.2:4b: Robot: moved page)
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{{NAVCONTENT_START}}
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We could substitute
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<center> [[Image:2_2_4b-1(2).gif]] </center>
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<math>u=x-\text{1}</math>, but we would then still have the problem of the second term,
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<math>\text{3}</math>
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in the denominator. Instead, we take out a factor
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<center> [[Image:2_2_4b-2(2).gif]] </center>
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<math>\text{3}</math>
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{{NAVCONTENT_STOP}}
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from the denominator,
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<math>\begin{align}
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& \int{\frac{\,dx}{\left( x-1 \right)^{2}+3}}=\int{\frac{\,dx}{3\left( \frac{1}{3}\left( x-1 \right)^{2}+1 \right)}} \\
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& =\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}} \\
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\end{align}</math>
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and move a factor
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<math>\frac{1}{3}</math>
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into the square
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<math>\left( x-1 \right)^{2}</math>,
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<math>\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}}=\frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}</math>
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Now, we substitute
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<math>u=\frac{x-1}{\sqrt{3}}</math>
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and get rid of all the problems at once:
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<math>\begin{align}
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& \frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}=\left\{ \begin{matrix}
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u=\frac{x-1}{\sqrt{3}} \\
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du=\frac{\,dx}{\sqrt{3}} \\
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\end{matrix} \right\} \\
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& =\frac{1}{3}\int{\frac{\sqrt{3}\,du}{u^{2}+1}}=\frac{\sqrt{3}}{3}\int{\frac{\,du}{u^{2}+1}} \\
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& =\frac{1}{\sqrt{3}}\arctan u+C \\
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& =\frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}}+C \\
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\end{align}</math>

Version vom 12:38, 21. Okt. 2008

We could substitute u=x−1, but we would then still have the problem of the second term, 3 in the denominator. Instead, we take out a factor 3 from the denominator,


dxx−12+3=dx331x−12+1=31dx31x−12+1

and move a factor 31 into the square x−12 ,


31dx31x−12+1=31dx3x−12+1 

Now, we substitute u=3x−1 and get rid of all the problems at once:


31dx3x−12+1=u=3x−1du=3dx=313duu2+1=33duu2+1=13arctanu+C=13arctan3x−1+C

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:4b“