Lösung 2.2:4b

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We could substitute
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We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,
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<math>u=x-\text{1}</math>, but we would then still have the problem of the second term,
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<math>\text{3}</math>
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in the denominator. Instead, we take out a factor
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<math>\text{3}</math>
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from the denominator,
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{{Displayed math||<math>\begin{align}
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\int \frac{dx}{(x-1)^2+3}
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&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]
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&= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1}
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\end{align}</math>}}
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<math>\begin{align}
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and move a factor <math>\tfrac{1}{3}</math> into the square <math>(x-1)^2</math>,
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& \int{\frac{\,dx}{\left( x-1 \right)^{2}+3}}=\int{\frac{\,dx}{3\left( \frac{1}{3}\left( x-1 \right)^{2}+1 \right)}} \\
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& =\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}} \\
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\end{align}</math>
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and move a factor
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{{Displayed math||<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}
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<math>\frac{1}{3}</math>
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into the square
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<math>\left( x-1 \right)^{2}</math>,
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Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,
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<math>\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}}=\frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}</math>
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{{Displayed math||<math>\begin{align}
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\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
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Now, we substitute
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&= \left\{\begin{align}
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<math>u=\frac{x-1}{\sqrt{3}}</math>
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u &= (x-1)/\!\sqrt{3}\\[5pt]
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and get rid of all the problems at once:
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du &= dx/\!\sqrt{3}
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\end{align}\right\}\\[5pt]
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&= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt]
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<math>\begin{align}
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&= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt]
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& \frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}=\left\{ \begin{matrix}
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&= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt]
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u=\frac{x-1}{\sqrt{3}} \\
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&= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.}
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du=\frac{\,dx}{\sqrt{3}} \\
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\end{align}</math>}}
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\end{matrix} \right\} \\
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& =\frac{1}{3}\int{\frac{\sqrt{3}\,du}{u^{2}+1}}=\frac{\sqrt{3}}{3}\int{\frac{\,du}{u^{2}+1}} \\
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& =\frac{1}{\sqrt{3}}\arctan u+C \\
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& =\frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}}+C \\
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\end{align}</math>
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Version vom 15:15, 28. Okt. 2008

We could substitute \displaystyle u=x-1, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,

\displaystyle \begin{align}

\int \frac{dx}{(x-1)^2+3} &= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt] &= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} \end{align}

and move a factor \displaystyle \tfrac{1}{3} into the square \displaystyle (x-1)^2,

\displaystyle \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}

Now, we substitute \displaystyle u = (x-1)/\!\sqrt{3} and get rid of all the problems at once,

\displaystyle \begin{align}

\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1} &= \left\{\begin{align} u &= (x-1)/\!\sqrt{3}\\[5pt] du &= dx/\!\sqrt{3} \end{align}\right\}\\[5pt] &= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt] &= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt] &= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt] &= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:4b“