Lösung 2.1:3d

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{{NAVCONTENT_START}}
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By dividing the two terms in the numerator by
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<center> [[Image:2_1_3d.gif]] </center>
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<math>x</math>, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand:
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& \int{\frac{x^{2}+1}{x}}\,dx=\int{\left( \frac{x^{2}}{x}+\frac{1}{x} \right)}\,dx \\
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& =\int{\left( x+x^{-1} \right)}\,dx \\
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& =\frac{x^{2}}{2}+\ln \left| x \right|+C \\
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\end{align}</math>
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where
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<math>C</math>
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is an arbitrary constant.
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NOTE: observe that
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<math>\frac{1}{x}</math>
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has a singularity at
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<math>x=0</math>, so the answers above are only primitive functions over intervals that do not contain
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<math>x=0</math>.

Version vom 13:57, 17. Okt. 2008

By dividing the two terms in the numerator by \displaystyle x, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand:


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


where \displaystyle C is an arbitrary constant.

NOTE: observe that \displaystyle \frac{1}{x} has a singularity at \displaystyle x=0, so the answers above are only primitive functions over intervals that do not contain \displaystyle x=0.

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