Lösung 2.2:4c

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The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
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{{Displayed math||<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}
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<math>\int{\frac{\,dx}{x^{2}+4x+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}-2^{2}+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}</math>
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We take out a factor 4 from the denominator
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We take out a factor
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<math>\text{4}</math>
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from the denominator
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<math>\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}\left( x+2 \right)^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}</math>
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{{Displayed math||<math>\int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}</math>}}
and rewrite the quadratic term as
and rewrite the quadratic term as
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{{Displayed math||<math>\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}</math>}}
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<math>\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}</math>
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If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise
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If we now substitute
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<math>u=\frac{x+2}{2}</math>, we obtain the integral in the exercise
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}=\left\{ \begin{matrix}
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\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
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u=\frac{x+2}{2} \\
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&= \left\{\begin{align}
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du=\frac{\,dx}{2} \\
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u &= (x+2)/2\\[5pt]
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\end{matrix} \right\} \\
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du &= dx/2
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& =\frac{1}{4}\int{\frac{\,2dx}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,dx}{u^{2}+1}} \\
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\end{align}\right\}\\[5pt]
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& =\frac{1}{2}\arctan u+C \\
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&= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt]
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& =\frac{1}{2}\arctan \frac{x+2}{2}+C \\
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&= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt]
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\end{align}</math>
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&= \frac{1}{2}\arctan u + C\\[5pt]
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&= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.}
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\end{align}</math>}}

Version vom 15:27, 28. Okt. 2008

The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,

\displaystyle \int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}

We take out a factor 4 from the denominator

\displaystyle \int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}

and rewrite the quadratic term as

\displaystyle \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}

If we now substitute \displaystyle u = (x+2)/2, we obtain the integral in the exercise

\displaystyle \begin{align}

\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1} &= \left\{\begin{align} u &= (x+2)/2\\[5pt] du &= dx/2 \end{align}\right\}\\[5pt] &= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt] &= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt] &= \frac{1}{2}\arctan u + C\\[5pt] &= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.} \end{align}

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