Lösung 2.2:4c

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The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
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<center> [[Image:2_2_4c-1(2).gif]] </center>
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<math>\int{\frac{\,dx}{x^{2}+4x+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}-2^{2}+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}</math>
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<center> [[Image:2_2_4c-2(2).gif]] </center>
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We take out a factor
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<math>\text{4}</math>
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from the denominator
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<math>\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}\left( x+2 \right)^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}</math>
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and rewrite the quadratic term as
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<math>\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}</math>
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If we now substitute
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<math>u=\frac{x+2}{2}</math>, we obtain the integral in the exercise
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<math>\begin{align}
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& \frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}=\left\{ \begin{matrix}
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u=\frac{x+2}{2} \\
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du=\frac{\,dx}{2} \\
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\end{matrix} \right\} \\
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& =\frac{1}{4}\int{\frac{\,2dx}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,dx}{u^{2}+1}} \\
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& =\frac{1}{2}\arctan u+C \\
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& =\frac{1}{2}\arctan \frac{x+2}{2}+C \\
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\end{align}</math>

Version vom 12:51, 21. Okt. 2008

The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,


\displaystyle \int{\frac{\,dx}{x^{2}+4x+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}-2^{2}+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}


We take out a factor \displaystyle \text{4} from the denominator


\displaystyle \int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}\left( x+2 \right)^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}


and rewrite the quadratic term as


\displaystyle \frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}


If we now substitute \displaystyle u=\frac{x+2}{2}, we obtain the integral in the exercise


\displaystyle \begin{align} & \frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}=\left\{ \begin{matrix} u=\frac{x+2}{2} \\ du=\frac{\,dx}{2} \\ \end{matrix} \right\} \\ & =\frac{1}{4}\int{\frac{\,2dx}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,dx}{u^{2}+1}} \\ & =\frac{1}{2}\arctan u+C \\ & =\frac{1}{2}\arctan \frac{x+2}{2}+C \\ \end{align}

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