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Lösung 2.2:4b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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,
-
{{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 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,

dx(x−1)2+3=dx331(x−1)2+1=31dx31(x−1)2+1

and move a factor 31 into the square (x−1)2,

31dx31(x−1)2+1=31dx3x−12+1.

Now, we substitute u=(x−1)3  and get rid of all the problems at once,

31dx3x−12+1=udu=(x−1)3=dx3=313duu2+1=33duu2+1=13arctanu+C=13arctan3x−1+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:4b“