Lösung 2.2:3d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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Observe that the derivative of the denominator is, for the most part, equal to the numerator,
Observe that the derivative of the denominator is, for the most part, equal to the numerator,
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{{Displayed math||<math>(x^2+2x+2)' = 2x+2 = 2(x+1)</math>}}
+
{{Abgesetzte Formel||<math>(x^2+2x+2)' = 2x+2 = 2(x+1)</math>}}
so we can rewrite the integral as
so we can rewrite the integral as
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{{Displayed math||<math>\int \frac{\tfrac{1}{2}}{x^2+2x+2}\cdot (x^2+2x+2)'\,dx\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\int \frac{\tfrac{1}{2}}{x^2+2x+2}\cdot (x^2+2x+2)'\,dx\,\textrm{.}</math>}}
The substitution <math>u=x^2+2x+2</math> will therefore simplify the integral considerably,
The substitution <math>u=x^2+2x+2</math> will therefore simplify the integral considerably,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int \frac{x+1}{x^2+2x+2}\,dx
\int \frac{x+1}{x^2+2x+2}\,dx
&= \left\{\begin{align}
&= \left\{\begin{align}
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Note: By completing the square
Note: By completing the square
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{{Displayed math||<math>x^2+2x+2 = (x+1)^2-1^2+2 = (x+1)^2+1</math>}}
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{{Abgesetzte Formel||<math>x^2+2x+2 = (x+1)^2-1^2+2 = (x+1)^2+1</math>}}
we see that <math>x^2+2x+2</math> is always greater than or equal to 1, so we can take away the absolute sign around the argument in <math>\ln</math> and answer with
we see that <math>x^2+2x+2</math> is always greater than or equal to 1, so we can take away the absolute sign around the argument in <math>\ln</math> and answer with
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{{Displayed math||<math>\frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}</math>}}

Version vom 13:02, 10. Mär. 2009

Observe that the derivative of the denominator is, for the most part, equal to the numerator,

\displaystyle (x^2+2x+2)' = 2x+2 = 2(x+1)

so we can rewrite the integral as

\displaystyle \int \frac{\tfrac{1}{2}}{x^2+2x+2}\cdot (x^2+2x+2)'\,dx\,\textrm{.}

The substitution \displaystyle u=x^2+2x+2 will therefore simplify the integral considerably,

\displaystyle \begin{align}

\int \frac{x+1}{x^2+2x+2}\,dx &= \left\{\begin{align} u &= x^2+2x+2\\[5pt] du &= (x^2+2x+2)'\,dx = 2(x+1)\,dx \end{align}\right\}\\[5pt] &= \frac{1}{2}\int \frac{du}{u}\\[5pt] &= \frac{1}{2}\ln |u| + C\\[5pt] &= \frac{1}{2}\ln |x^2+2x+2| + C\,\textrm{.} \end{align}


Note: By completing the square

\displaystyle x^2+2x+2 = (x+1)^2-1^2+2 = (x+1)^2+1

we see that \displaystyle x^2+2x+2 is always greater than or equal to 1, so we can take away the absolute sign around the argument in \displaystyle \ln and answer with

\displaystyle \frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3d“