Lösung 2.3:2d

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We shall solve the exercise in two different ways.
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Method 1 (partial integration)
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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


\displaystyle 1\centerdot \ln x


We integrate the factor \displaystyle \text{1} and differentiate \displaystyle \ln x,


\displaystyle \begin{align} & \int{1\centerdot \ln x\,dx}=x\centerdot \ln x-\int{x\centerdot \frac{1}{x}\,dx} \\ & =x\centerdot \ln x-\int{1\,dx} \\ & =x\centerdot \ln x-x+C \\ \end{align}


Method 2 (substitution)

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


\displaystyle du=\left( \ln x \right)^{\prime }\,dx=\frac{1}{x}\,dx


and because \displaystyle u=\text{ln }x, then \displaystyle x=e^{u} and we have that


\displaystyle du=\frac{1}{e^{u}\,}\,dx\quad \Leftrightarrow \quad dx=e^{u}\,du


Thus, the substitution becomes


\displaystyle \begin{align} & \int{\ln x\,dx}=\left\{ \begin{matrix} u=\text{ln }x \\ dx=e^{u}\,du \\ \end{matrix} \right\} \\ & =\int{ue^{u}\,du} \\ \end{align}


Now, we carry out a partial integration


\displaystyle \begin{align} & \int{ue^{u}\,du}=u\centerdot e^{u}-\int{1\centerdot e^{u}\,du} \\ & =u\centerdot e^{u}-\int{e^{u}\,du} \\ & =u\centerdot e^{u}-e^{u}+C \\ & =\left( u-1 \right)e^{u}+C \\ \end{align}


and the answer becomes


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

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