Lösung 2.2:3d

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Observe that the derivative of the denominator is, for the most part, equal to the numerator,
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<center> [[Image:2_2_3d.gif]] </center>
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<math>\left( x^{2}+2x+2 \right)^{\prime }=2x+2=2\left( x+1 \right)</math>
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so we can rewrite the integral as
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<math>\int{\frac{\frac{1}{2}}{x^{2}+2x+2}}\centerdot \left( x^{2}+2x+2 \right)^{\prime }\,dx</math>
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The substitution
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<math>u=x^{2}+2x+2</math>
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will therefore simplify the integral considerably:
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<math>\begin{align}
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& \int{\frac{x+1}{x^{2}+2x+2}}\,dx=\left\{ \begin{matrix}
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u=x^{2}+2x+2 \\
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du=\left( x^{2}+2x+2 \right)^{\prime }\,dx=2\left( x+1 \right)\,dx \\
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\end{matrix} \right\} \\
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& =\frac{1}{2}\int{\frac{\,du}{u}}=\frac{1}{2}\ln \left| u \right|+C \\
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& =\frac{1}{2}\ln \left| x^{2}+2x+2 \right|+C \\
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\end{align}</math>
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NOTE: By completing the square
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<math>x^{2}+2x+2=\left( x+1 \right)^{2}-1^{2}+2=\left( x+1 \right)^{2}+1</math>
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we see that
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<math>x^{2}+2x+2</math>
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is always greater than or equal to
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<math>\text{1}</math>, so we can take away the absolute sign around the argument in
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<math>\text{ln}</math>
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and answer with
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<math>\frac{1}{2}\ln \left( x^{2}+2x+2 \right)+C</math>

Version vom 10:51, 21. Okt. 2008

Observe that the derivative of the denominator is, for the most part, equal to the numerator,


\displaystyle \left( x^{2}+2x+2 \right)^{\prime }=2x+2=2\left( x+1 \right)


so we can rewrite the integral as


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


The substitution \displaystyle u=x^{2}+2x+2 will therefore simplify the integral considerably:


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

NOTE: By completing the square


\displaystyle x^{2}+2x+2=\left( x+1 \right)^{2}-1^{2}+2=\left( x+1 \right)^{2}+1


we see that \displaystyle x^{2}+2x+2 is always greater than or equal to \displaystyle \text{1}, so we can take away the absolute sign around the argument in \displaystyle \text{ln} and answer with


\displaystyle \frac{1}{2}\ln \left( x^{2}+2x+2 \right)+C

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3d“