Lösung 2.2:4a
Aus Online Mathematik Brückenkurs 2
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Version vom 10:24, 11. 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} |
