Lösung 2.2:4b

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,
We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int \frac{dx}{(x-1)^2+3}
\int \frac{dx}{(x-1)^2+3}
&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]
&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]
Zeile 9: Zeile 9:
and move a factor <math>\tfrac{1}{3}</math> into the square <math>(x-1)^2</math>,
and move a factor <math>\tfrac{1}{3}</math> into the square <math>(x-1)^2</math>,
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{{Displayed math||<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}
Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,
Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
&= \left\{\begin{align}
&= \left\{\begin{align}

Version vom 13:02, 10. Mär. 2009

We could substitute \displaystyle u=x-1, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,

\displaystyle \begin{align}

\int \frac{dx}{(x-1)^2+3} &= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt] &= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} \end{align}

and move a factor \displaystyle \tfrac{1}{3} into the square \displaystyle (x-1)^2,

\displaystyle \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}

Now, we substitute \displaystyle u = (x-1)/\!\sqrt{3} and get rid of all the problems at once,

\displaystyle \begin{align}

\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1} &= \left\{\begin{align} u &= (x-1)/\!\sqrt{3}\\[5pt] du &= dx/\!\sqrt{3} \end{align}\right\}\\[5pt] &= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt] &= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt] &= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt] &= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.} \end{align}

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