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Aus Online Mathematik Brückenkurs 2

We could substitute 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,

dx(x−1)2+3=dx331(x−1)2+1=31dx31(x−1)2+1

and move a factor 31 into the square (x−1)2,

31dx31(x−1)2+1=31dx3x−12+1.

Now, we substitute u=(x−1)3  and get rid of all the problems at once,

31dx3x−12+1=udu=(x−1)3=dx3=313duu2+1=33duu2+1=13arctanu+C=13arctan3x−1+C.