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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=dxx+22−22+8=dxx+22+4 

We take out a factor 4 from the denominator

dxx+22+4=dx441x+22+1=41dx41x+22+1 

and rewrite the quadratic term as

41dx41x+22+1=41dx2x+22+1 

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

41dx2x+22+1=u=2x+2du=2dx=412dxu2+1=21dxu2+1=21arctanu+C=21arctan2x+2+C