Lösung 2.1:3c

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If we multiply the factors in the integrand together and use the power laws,
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<center> [[Image:2_1_3c.gif]] </center>
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<math>\begin{align}
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& \int{e^{2x}}\left( e^{x}+1 \right)\,dx=\int{\left( e^{2x}e^{x}+e^{2x} \right)}\,dx \\
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& =\int{\left( e^{2x+x}+e^{2x} \right)}\,dx=\int{\left( e^{3x}+e^{2x} \right)}\,dx \\
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\end{align}</math>
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we obtain a standard integral with two terms of the type
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<math>e^{ax}</math>, where
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<math>a</math>
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is a constant. The indefinite integral is therefore
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<math>\int{\left( e^{3x}+e^{2x} \right)}\,dx=\frac{e^{3x}}{3}+\frac{e^{2x}}{2}+C</math>
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where
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<math>C</math>
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is an arbitrary constant.

Version vom 13:48, 17. Okt. 2008

If we multiply the factors in the integrand together and use the power laws,


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

we obtain a standard integral with two terms of the type \displaystyle e^{ax}, where \displaystyle a is a constant. The indefinite integral is therefore


\displaystyle \int{\left( e^{3x}+e^{2x} \right)}\,dx=\frac{e^{3x}}{3}+\frac{e^{2x}}{2}+C

where \displaystyle C is an arbitrary constant.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:3c“