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