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,

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=dx441(x+2)2+1=41dx41(x+2)2+1

and rewrite the quadratic term as

41dx41(x+2)2+1=41dx2x+22+1.

If we now substitute u=(x+2)2, we obtain the integral in the exercise

41dx2x+22+1=udu=(x+2)2=dx2=412duu2+1=21duu2+1=21arctanu+C=21arctan2x+2+C.