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Lösung 2.2:4a

Aus Online Mathematik Brückenkurs 2

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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 x2+4 instead of x2+1, but if we factor out the 4 from the denominator,

dxx2+4=dx441x2+1=41dx41x2+1, 

we obtain the correct second term in the denominator. On the other hand, there is no longer x2 but 41x2, although we can get around this by substituting u=21x,

41dx41x2+1=41dx(x2)2+1=udu=x2=21dx=412duu2+1=21duu2+1=21arctanu+C=21arctanx2+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:4a“