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Lösung 2.2:4c
Aus Online Mathematik Brückenkurs 2
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Version vom 10:25, 11. Mär. 2009
The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
 dxx2+4x+8=dx(x+2)2−22+8=dx(x+2)2+4. |
We take out a factor 4 from the denominator
dx(x+2)2+4=dx4 41(x+2)2+1 =41dx41(x+2)2+1 |
and rewrite the quadratic term as
dx41(x+2)2+1=41dx 2x+2 2+1. |
If we now substitute 
2
dx2x+22+1= udu=(x+2)2=dx2 =412duu2+1=21duu2+1=21arctanu+C=21arctan2x+2+C. |
