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Lösung 2.3:2d

Aus Online Mathematik Brückenkurs 2

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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)
 +
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>}}
 +
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
-
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
+
-
<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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-
 
+
-
 
+
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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>
+
-
and we have that
+
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+
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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
 +
{{Displayed math||<math>\begin{align}
 +
\int \ln x\,dx
 +
= \left\{\begin{align}
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u &= \ln x\\[5pt]
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dx &= e^u\,du
 +
\end{align}\right\}
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= \int ue^u\,du\,\textrm{.}
 +
\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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+
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-
Now, we carry out a partial integration
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-
 
+
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+
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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}
 +
\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}</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>
+

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

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=lnx. 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=(lnx)dx=x1dx

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

du=1eudxdx=eudu.

Thus, the substitution becomes

lnxdx=udx=lnx=eudu=ueudu. 

Now, we carry out an integration by parts,

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

and the answer becomes

lnxdx=(lnx−1)elnx+C=(lnx−1)x+C. 
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:2d“