Processing Math: Done
Lösung 2.1:3d
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 2.1:3d moved to Solution 2.1:3d: Robot: moved page) |
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- | {{ | + | 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>\begin{align} | ||
+ | & \int{\frac{x^{2}+1}{x}}\,dx=\int{\left( \frac{x^{2}}{x}+\frac{1}{x} \right)}\,dx \\ | ||
+ | & =\int{\left( x+x^{-1} \right)}\,dx \\ | ||
+ | & =\frac{x^{2}}{2}+\ln \left| x \right|+C \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | |||
+ | where | ||
+ | <math>C</math> | ||
+ | is an arbitrary constant. | ||
+ | |||
+ | NOTE: observe that | ||
+ | <math>\frac{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>. |
Version vom 13:57, 17. Okt. 2008
By dividing the two terms in the numerator by
xx2+1dx=
xx2+x1
dx=
x+x−1
dx=2x2+ln
x
+C
where
NOTE: observe that