Lösung 2.3:2d

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We shall solve the exercise in two different ways.
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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'''Method 1''' (integration by parts)
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At first sight, integration by parts 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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{{Displayed math||<math>1\centerdot \ln x\,\textrm{.}</math>}}
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We integrate the factor <math>1</math> and differentiate <math>\ln x</math>,
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We integrate the factor
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{{Displayed math||<math>\begin{align}
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<math>\text{1}</math>
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\int 1\cdot\ln x\,dx
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and differentiate
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&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]
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<math>\ln x</math>,
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&= x\cdot\ln x - \int 1\,dx\\[5pt]
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&= x\cdot\ln x - x + C\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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'''Method 2''' (substitution)
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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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It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression <math>u=\ln x\,</math>. The problem we encounter is how we should handle the change from <math>dx</math> to <math>du</math>. With this substitution, the relation between <math>dx</math> and <math>du</math> becomes
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Method 2 (substitution)
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{{Displayed math||<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}
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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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and because <math>u = \ln x</math>, then <math>x=e^u</math> and we have that
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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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{{Displayed math||<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}
Thus, the substitution becomes
Thus, the substitution becomes
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{{Displayed math||<math>\begin{align}
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\int \ln x\,dx
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= \left\{\begin{align}
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u &= \ln x\\[5pt]
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dx &= e^u\,du
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\end{align}\right\}
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= \int ue^u\,du\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Now, we carry out an integration by parts,
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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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{{Displayed math||<math>\begin{align}
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\int u\cdot e^u\,du
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&= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt]
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&= ue^u - \int e^u\,du\\[5pt]
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&= ue^u - e^u + C\\[5pt]
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&= (u-1)e^u + C\,,
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\end{align}</math>}}
and the answer becomes
and the answer becomes
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{{Displayed math||<math>\begin{align}
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\int \ln x\,dx
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<math>\begin{align}
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&= (\ln x-1)e^{\ln x} + C\\[5pt]
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& \int{\ln x\,dx}=\left( \ln x-1 \right)e^{\ln x}+C \\
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&= (\ln x-1)x + C\,\textrm{.}
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& =\left( \ln x-1 \right)x+C \\
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\end{align}</math>}}
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\end{align}</math>
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Version vom 09:13, 29. Okt. 2008

We shall solve the exercise in two different ways.


Method 1 (integration by parts)

At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product

\displaystyle 1\centerdot \ln x\,\textrm{.}

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

\displaystyle \begin{align}

\int 1\cdot\ln x\,dx &= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt] &= x\cdot\ln x - \int 1\,dx\\[5pt] &= x\cdot\ln x - x + C\,\textrm{.} \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=\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 = (\ln x)'\,dx = \frac{1}{x}\,dx

and because \displaystyle u = \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\,\textrm{.}

Thus, the substitution becomes

\displaystyle \begin{align}

\int \ln x\,dx = \left\{\begin{align} u &= \ln x\\[5pt] dx &= e^u\,du \end{align}\right\} = \int ue^u\,du\,\textrm{.} \end{align}

Now, we carry out an integration by parts,

\displaystyle \begin{align}

\int u\cdot e^u\,du &= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt] &= ue^u - \int e^u\,du\\[5pt] &= ue^u - e^u + C\\[5pt] &= (u-1)e^u + C\,, \end{align}

and the answer becomes

\displaystyle \begin{align}

\int \ln x\,dx &= (\ln x-1)e^{\ln x} + C\\[5pt] &= (\ln x-1)x + C\,\textrm{.} \end{align}

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