Lösung 2.2:4a

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What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,
What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,
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{{Displayed math||<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}
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{{Abgesetzte Formel||<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}
we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,
we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}

Version vom 13:02, 10. Mär. 2009

What makes our integral differ from that in the exercise´s text is that there is a term \displaystyle x^2+4 instead of \displaystyle x^2+1, but if we factor out the 4 from the denominator,

\displaystyle \int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}

we obtain the correct second term in the denominator. On the other hand, there is no longer \displaystyle x^2 but \displaystyle \tfrac{1}{4}x^2, although we can get around this by substituting \displaystyle u=\tfrac{1}{2}x,

\displaystyle \begin{align}

\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1} &= \frac{1}{4}\int \frac{dx}{(x/2)^2+1} = \left\{\begin{align} u &= x/2\\[5pt] du &= \tfrac{1}{2}\,dx \end{align}\right\}\\[5pt] &= \frac{1}{4}\int \frac{2\,du}{u^2+1} = \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt] &= \frac{1}{2}\arctan u + C = \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.} \end{align}

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