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Lösung 2.1:3c

Aus Online Mathematik Brückenkurs 2

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If we multiply the factors in the integrand together and use the power laws,
If we multiply the factors in the integrand together and use the power laws,
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{{Displayed math||<math>\begin{align}
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\int e^{2x}\bigl(e^x+1\bigr)\,dx
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&= \int\bigl(e^{2x}e^{x} + e^{2x}\bigr)\,dx\\[5pt]
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&= \int\bigl(e^{2x+x} + e^{2x}\bigr)\,dx\\[5pt]
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&= \int{\bigl(e^{3x} + e^{2x}\bigr)}\,dx\,,
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\end{align}</math>}}
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<math>\begin{align}
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we obtain a standard integral with two terms of the type <math>e^{ax}</math>, where
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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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<math>a</math> is a constant. The indefinite integral is therefore
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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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{{Displayed math||<math>\int \bigl(e^{3x}+e^{2x}\bigr)\,dx = \frac{e^{3x}}{3} + \frac{e^{2x}}{2} + C\,,</math>}}
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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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-
 
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where <math>C</math> is an arbitrary constant.
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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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-
 
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where
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<math>C</math>
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is an arbitrary constant.
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Version vom 13:19, 21. Okt. 2008

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

e2xex+1dx=e2xex+e2xdx=e2x+x+e2xdx=e3x+e2xdx

we obtain a standard integral with two terms of the type eax, where a is a constant. The indefinite integral is therefore

e3x+e2xdx=3e3x+2e2x+C 

where C is an arbitrary constant.

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