Solution 2.2:4d

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Current revision (15:36, 28 October 2008) (edit) (undo)
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The integral can be simplified by a so-called polynomial division. We add and take away
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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
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<math>\text{1}</math>
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in the numerator and can thus eliminate the
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<math>x^{2}</math>
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-term from the numerator
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<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}</math>
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{{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>}}
Thus, we have
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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<math>\int{\frac{x^{2}}{x^{2}+1}\,dx=\int{\left( 1-\frac{1}{x^{2}+1} \right)}}\,dx=x-\arctan x+C</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{.}
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