How can I integrate ∫ 1 2 x + 2 Method 1 ∫ 1 2 x + 2 = 1 2 ∫ 1 x + 1 = 1 2 l n ( x + 1 ) + cMethod 2 ∫ 1 2 x + 2 = 1 2 ∫ 2 2 x + 2 = 1 2 l n ( 2 x + 2 ) + cWolfram Alpha suggests the first method yields the correct answer, but why do the two methods produce different answers?

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How can I integrate   ∫            1              2        x        +        2            Method 1  ∫            1              2        x        +        2              =      1    2    ∫            1              x        +        1              =      1    2    l  n  (  x  +  1  )  +  cMethod 2  ∫            1              2        x        +        2              =      1    2    ∫            2              2        x        +        2              =      1    2    l  n  (  2  x  +  2  )  +  cWolfram Alpha suggests the first method yields the correct answer, but why do the two methods produce different answers?

bideanbarrenaf5 · Accepted Answer

You can do this  ∫            1              2        x        +        2              d  x  =      1    2    ∫                    d        x                    x        +        1              =      1    2    ∫                    d        u            u        =      1    2    ln  ⁡      u    +  cwhere   u  =  x  +  1,   d  u  =  d  x. You know that  ∫                    d        u            u        =  ln  ⁡      u    +  c

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