From Förberedande kurs i matematik 2

Revision as of 14:46, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:4d moved to Solution 2.2:4d: Robot: moved page) ← Previous diff		Current revision (15:36, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The integral can be simplified by a so-called polynomial division. We add and take away 1 in the numerator and can thus eliminate the <math>x^2</math>-term from the numerator
-	<center> [[Image:2_2_4d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\frac{x^2}{x^{2}+1} = \frac{x^2+1-1}{x^2+1} = \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} = 1-\frac{1}{x^2+1}\,\textrm{.}</math>}}
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		+	Thus, we have
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		+	{{Displayed math||<math>\int\frac{x^2}{x^2+1}\,dx = \int\Bigl(1-\frac{1}{x^2+1} \Bigr)\,dx = x-\arctan x+C\,\textrm{.}</math>}}

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Current revision

The integral can be simplified by a so-called polynomial division. We add and take away 1 in the numerator and can thus eliminate the \displaystyle x^2-term from the numerator

\displaystyle \frac{x^2}{x^{2}+1} = \frac{x^2+1-1}{x^2+1} = \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} = 1-\frac{1}{x^2+1}\,\textrm{.}

Thus, we have

\displaystyle \int\frac{x^2}{x^2+1}\,dx = \int\Bigl(1-\frac{1}{x^2+1} \Bigr)\,dx = x-\arctan x+C\,\textrm{.}