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Lösung 2.3:2d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.3:2d moved to Solution 2.3:2d: Robot: moved page)
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We shall solve the exercise in two different ways.
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<center> [[Image:2_3_2d-1(2).gif]] </center>
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Method 1 (partial integration)
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<center> [[Image:2_3_2d-2(2).gif]] </center>
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As first sight, partial integration seems impossible, but the trick is to see the integrand as the product
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<math>1\centerdot \ln x</math>
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We integrate the factor
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<math>\text{1}</math>
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and differentiate
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<math>\ln x</math>,
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<math>\begin{align}
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& \int{1\centerdot \ln x\,dx}=x\centerdot \ln x-\int{x\centerdot \frac{1}{x}\,dx} \\
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& =x\centerdot \ln x-\int{1\,dx} \\
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& =x\centerdot \ln x-x+C \\
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\end{align}</math>
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Method 2 (substitution)
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It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression
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<math>u=\text{ln }x</math>. The problem we encounter is how we should handle the change from
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<math>dx</math>
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to
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<math>du</math>. With this substitution, the relation between
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<math>dx</math>
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and
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<math>du</math>
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becomes
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<math>du=\left( \ln x \right)^{\prime }\,dx=\frac{1}{x}\,dx</math>
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and because
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<math>u=\text{ln }x</math>, then
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<math>x=e^{u}</math>
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and we have that
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<math>du=\frac{1}{e^{u}\,}\,dx\quad \Leftrightarrow \quad dx=e^{u}\,du</math>
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Thus, the substitution becomes
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<math>\begin{align}
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& \int{\ln x\,dx}=\left\{ \begin{matrix}
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u=\text{ln }x \\
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dx=e^{u}\,du \\
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\end{matrix} \right\} \\
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& =\int{ue^{u}\,du} \\
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\end{align}</math>
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Now, we carry out a partial integration
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<math>\begin{align}
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& \int{ue^{u}\,du}=u\centerdot e^{u}-\int{1\centerdot e^{u}\,du} \\
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& =u\centerdot e^{u}-\int{e^{u}\,du} \\
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& =u\centerdot e^{u}-e^{u}+C \\
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& =\left( u-1 \right)e^{u}+C \\
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\end{align}</math>
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and the answer becomes
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<math>\begin{align}
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& \int{\ln x\,dx}=\left( \ln x-1 \right)e^{\ln x}+C \\
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& =\left( \ln x-1 \right)x+C \\
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\end{align}</math>

Version vom 14:32, 22. Okt. 2008

We shall solve the exercise in two different ways.

Method 1 (partial integration)

As first sight, partial integration seems impossible, but the trick is to see the integrand as the product


1lnx


We integrate the factor 1 and differentiate lnx,


1lnxdx=xlnx−xx1dx=xlnx−1dx=xlnx−x+C


Method 2 (substitution)

It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression u=ln x. The problem we encounter is how we should handle the change from dx to du. With this substitution, the relation between dx and du becomes


du=lnxdx=x1dx 


and because u=ln x, then x=eu and we have that


du=1eudxdx=eudu


Thus, the substitution becomes


lnxdx=u=ln xdx=eudu=ueudu 


Now, we carry out a partial integration


ueudu=ueu−1eudu=ueu−eudu=ueu−eu+C=u−1eu+C


and the answer becomes


lnxdx=lnx−1elnx+C=lnx−1x+C 

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