Lösung 2.2:4b

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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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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 \displaystyle u=x-\text{1}, but we would then still have the problem of the second term, \displaystyle \text{3} in the denominator. Instead, we take out a factor \displaystyle \text{3} from the denominator,


\displaystyle \begin{align} & \int{\frac{\,dx}{\left( x-1 \right)^{2}+3}}=\int{\frac{\,dx}{3\left( \frac{1}{3}\left( x-1 \right)^{2}+1 \right)}} \\ & =\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}} \\ \end{align}

and move a factor \displaystyle \frac{1}{3} into the square \displaystyle \left( x-1 \right)^{2},


\displaystyle \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}}

Now, we substitute \displaystyle u=\frac{x-1}{\sqrt{3}} and get rid of all the problems at once:


\displaystyle \begin{align} & \frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}=\left\{ \begin{matrix} u=\frac{x-1}{\sqrt{3}} \\ du=\frac{\,dx}{\sqrt{3}} \\ \end{matrix} \right\} \\ & =\frac{1}{3}\int{\frac{\sqrt{3}\,du}{u^{2}+1}}=\frac{\sqrt{3}}{3}\int{\frac{\,du}{u^{2}+1}} \\ & =\frac{1}{\sqrt{3}}\arctan u+C \\ & =\frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}}+C \\ \end{align}

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