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
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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,
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<math>x^{2}+4</math>
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instead of
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<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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<math>\int{\frac{\,dx}{x^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}x^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}x^{2}+1}}</math>,
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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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we obtain the correct second term in the denominator. On the other hand, there is no longer
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{{Displayed math||<math>\begin{align}
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<math>x^{2}</math>
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\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
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but
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&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
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<math>\frac{1}{4}x^{2}</math>, although we can get around this by substituting
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= \left\{\begin{align}
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<math>u=\frac{1}{2}x</math>,
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u &= x/2\\[5pt]
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du &= \tfrac{1}{2}\,dx
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\end{align}\right\}\\[5pt]
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<math>\begin{align}
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&= \frac{1}{4}\int \frac{2\,du}{u^2+1}
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& \frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}x^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x}{2} \right)^{2}+1}}=\left\{ \begin{matrix}
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= \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt]
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u=\frac{1}{2}x \\
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&= \frac{1}{2}\arctan u + C
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du=\frac{1}{2}\,dx \\
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= \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.}
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\end{matrix} \right\} \\
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\end{align}</math>}}
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& =\frac{1}{4}\int{\frac{2\,du}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,du}{u^{2}+1}} \\
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& =\frac{1}{2}\arctan u+C=\frac{1}{2}\arctan \frac{x}{2}+C \\
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\end{align}</math>
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Version vom 15:02, 28. Okt. 2008

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“