Lösung 2.1:3d
Aus Online Mathematik Brückenkurs 2
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| - | By dividing the two terms in the numerator by | + | By dividing the two terms in the numerator by <math>x</math>, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand, |
| - | <math>x</math>, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand | + | |
| + | {{Displayed math||<math>\begin{align} | ||
| + | \int \frac{x^{2}+1}{x}\,dx | ||
| + | &= \int \Bigl(\frac{x^2}{x} + \frac{1}{x}\Bigr)\,dx\\[5pt] | ||
| + | &= \int \bigl(x+x^{-1}\bigr)\,dx\\[5pt] | ||
| + | &= \frac{x^2}{2} + \ln |x| + C\,, | ||
| + | \end{align}</math>}} | ||
| - | <math> | + | where <math>C</math> is an arbitrary constant. |
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| - | + | Note: Observe that <math>1/x</math> has a singularity at <math>x=0</math>, so the answers above are only primitive functions over intervals that do not contain <math>x=0\,</math>. | |
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| - | <math> | + | |
| - | has a singularity at | + | |
| - | <math>x=0</math>, so the answers above are only primitive functions over intervals that do not contain | + | |
| - | <math>x=0</math>. | + | |
Version vom 13:23, 21. Okt. 2008
By dividing the two terms in the numerator by \displaystyle x, we can simplify each term to a form which makes it possible simply to write down the primitive functions of the integrand,
| \displaystyle \begin{align}
\int \frac{x^{2}+1}{x}\,dx &= \int \Bigl(\frac{x^2}{x} + \frac{1}{x}\Bigr)\,dx\\[5pt] &= \int \bigl(x+x^{-1}\bigr)\,dx\\[5pt] &= \frac{x^2}{2} + \ln |x| + C\,, \end{align} |
where \displaystyle C is an arbitrary constant.
Note: Observe that \displaystyle 1/x has a singularity at \displaystyle x=0, so the answers above are only primitive functions over intervals that do not contain \displaystyle x=0\,.
