Lösung 2.2:3d

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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>}}
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<math>\left( x^{2}+2x+2 \right)^{\prime }=2x+2=2\left( x+1 \right)</math>
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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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<math>\int{\frac{\frac{1}{2}}{x^{2}+2x+2}}\centerdot \left( x^{2}+2x+2 \right)^{\prime }\,dx</math>
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The substitution <math>u=x^2+2x+2</math> will therefore simplify the integral considerably,
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The substitution
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<math>u=x^{2}+2x+2</math>
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will therefore simplify the integral considerably:
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<math>\begin{align}
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& \int{\frac{x+1}{x^{2}+2x+2}}\,dx=\left\{ \begin{matrix}
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u=x^{2}+2x+2 \\
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du=\left( x^{2}+2x+2 \right)^{\prime }\,dx=2\left( x+1 \right)\,dx \\
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\end{matrix} \right\} \\
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& =\frac{1}{2}\int{\frac{\,du}{u}}=\frac{1}{2}\ln \left| u \right|+C \\
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& =\frac{1}{2}\ln \left| x^{2}+2x+2 \right|+C \\
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\end{align}</math>
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NOTE: By completing the square
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{{Displayed math||<math>\begin{align}
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\int \frac{x+1}{x^2+2x+2}\,dx
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&= \left\{\begin{align}
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u &= x^2+2x+2\\[5pt]
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du &= (x^2+2x+2)'\,dx = 2(x+1)\,dx
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\end{align}\right\}\\[5pt]
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&= \frac{1}{2}\int \frac{du}{u}\\[5pt]
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&= \frac{1}{2}\ln |u| + C\\[5pt]
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&= \frac{1}{2}\ln |x^2+2x+2| + C\,\textrm{.}
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\end{align}</math>}}
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<math>x^{2}+2x+2=\left( x+1 \right)^{2}-1^{2}+2=\left( x+1 \right)^{2}+1</math>
 
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Note: By completing the square
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we see that
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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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<math>x^{2}+2x+2</math>
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is always greater than or equal to
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<math>\text{1}</math>, so we can take away the absolute sign around the argument in
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<math>\text{ln}</math>
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and answer with
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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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<math>\frac{1}{2}\ln \left( x^{2}+2x+2 \right)+C</math>
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{{Displayed math||<math>\frac{1}{2}\ln (x^2+2x+2) + C\,\textrm{.}</math>}}

Version vom 14:34, 28. Okt. 2008

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“