Lösung 2.2:4b
Aus Online Mathematik Brückenkurs 2
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} |
