Lösung 2.3:2d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product
At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product
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{{Displayed math||<math>1\centerdot \ln x\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>1\centerdot \ln x\,\textrm{.}</math>}}
We integrate the factor <math>1</math> and differentiate <math>\ln x</math>,
We integrate the factor <math>1</math> and differentiate <math>\ln x</math>,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int 1\cdot\ln x\,dx
\int 1\cdot\ln x\,dx
&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]
&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]
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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
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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{{Displayed math||<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}
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{{Abgesetzte Formel||<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}
and because <math>u = \ln x</math>, then <math>x=e^u</math> and we have that
and because <math>u = \ln x</math>, then <math>x=e^u</math> and we have that
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{{Displayed math||<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<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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{{Abgesetzte Formel||<math>\begin{align}
\int \ln x\,dx
\int \ln x\,dx
= \left\{\begin{align}
= \left\{\begin{align}
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Now, we carry out an integration by parts,
Now, we carry out an integration by parts,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int u\cdot e^u\,du
\int u\cdot e^u\,du
&= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt]
&= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt]
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and the answer becomes
and the answer becomes
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int \ln x\,dx
\int \ln x\,dx
&= (\ln x-1)e^{\ln x} + C\\[5pt]
&= (\ln x-1)e^{\ln x} + C\\[5pt]
&= (\ln x-1)x + C\,\textrm{.}
&= (\ln x-1)x + C\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 13:04, 10. Mär. 2009

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“