Lösung 2.2:3e

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If we differentiate the denominator in the integrand
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<center> [[Image:2_2_3e.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\left( x^{2}+1 \right)^{\prime }</math>
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we obtain almost the same expression as in the numerator; there is a constant
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<math>\text{2}</math>
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which is different. We therefore rewrite the numerator as
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<math>3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\centerdot \left( x^{2}+1 \right)^{\prime }</math>,
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so the integral can be written as
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<math>\int{\frac{\frac{3}{2}}{x^{2}+1}}\centerdot \left( x^{2}+1 \right)^{\prime }\,dx</math>,
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and we see that the substitution
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<math>u=x^{2}+1</math>
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can be used to simplify the integral:
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<math>\begin{align}
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& \int{\frac{3x}{x^{2}+1}}\,dx=\left\{ \begin{matrix}
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u=x^{2}+1 \\
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du=\left( x^{2}+1 \right)^{\prime }\,dx=2x\,dx \\
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\end{matrix} \right\} \\
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& =\frac{3}{2}\int{\frac{\,du}{u}}=\frac{3}{2}\ln \left| u \right|+C \\
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& =\frac{3}{2}\ln \left| x^{2}+1 \right|+C \\
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& =\frac{3}{2}\ln \left( x^{2}+1 \right)+C \\
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\end{align}</math>
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In the last step, we take away the absolute sign around the argument in
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<math>\ln </math>, because
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<math>x^{2}+1</math>
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is always greater than or equal to
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<math>\text{1}</math>.

Version vom 11:16, 21. Okt. 2008

If we differentiate the denominator in the integrand


\displaystyle 3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\left( x^{2}+1 \right)^{\prime }


we obtain almost the same expression as in the numerator; there is a constant \displaystyle \text{2} which is different. We therefore rewrite the numerator as


\displaystyle 3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\centerdot \left( x^{2}+1 \right)^{\prime },

so the integral can be written as


\displaystyle \int{\frac{\frac{3}{2}}{x^{2}+1}}\centerdot \left( x^{2}+1 \right)^{\prime }\,dx,

and we see that the substitution \displaystyle u=x^{2}+1 can be used to simplify the integral:


\displaystyle \begin{align} & \int{\frac{3x}{x^{2}+1}}\,dx=\left\{ \begin{matrix} u=x^{2}+1 \\ du=\left( x^{2}+1 \right)^{\prime }\,dx=2x\,dx \\ \end{matrix} \right\} \\ & =\frac{3}{2}\int{\frac{\,du}{u}}=\frac{3}{2}\ln \left| u \right|+C \\ & =\frac{3}{2}\ln \left| x^{2}+1 \right|+C \\ & =\frac{3}{2}\ln \left( x^{2}+1 \right)+C \\ \end{align}


In the last step, we take away the absolute sign around the argument in \displaystyle \ln , because \displaystyle x^{2}+1 is always greater than or equal to \displaystyle \text{1}.

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