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Lösung 2.2:3e

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.2:3e moved to Solution 2.2:3e: Robot: moved page)
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{{NAVCONTENT_START}}
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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


3x=232x=23x2+1 


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


3x=232x=23x2+1 ,

so the integral can be written as


23x2+1x2+1dx ,

and we see that the substitution u=x2+1 can be used to simplify the integral:


3xx2+1dx=u=x2+1du=x2+1dx=2xdx=23udu=23lnu+C=23lnx2+1+C=23lnx2+1+C


In the last step, we take away the absolute sign around the argument in ln, because x2+1 is always greater than or equal to 1.

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