Finding the antiderivativeLet ∫ 1 + 1 x 2 x 2 − 1 + 1 x 2 d xI've been told use the substitution: t = x − 1 x .But how to apply it on the integral?

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Finding the antiderivativeLet   ∫                    1        +                  1                                                    x                2                                                                                      x            2                          −        1        +                  1                                                    x                2                                                          d  xI&#039;ve been told use the substitution:   t  =                    x        −        1            x      .But how to apply it on the integral?

Astok3mpd · Accepted Answer

Step 1      x    2    −  1  +            1              x        2              =            (      x      −                        1          x                    )        2    +  1Let       (    x    −                  1        x              )    =  z    ⟹    1  +  1      /        x    2    d  x  =  d  zStep 2So integral becomes   ∫                    d        z                              z          2                +        1              =  arctan  ⁡  (  z  )  =  arctan  ⁡      (    x    −                  1        x              )    +  C

Defensorentx9 · Accepted Answer

Step 1      x    2    −  1  +      1          x      2        =            (      x      −              1        x            )        2    +  1and since             (      x      −              1        x            )        ′    =  1  +      1          x      2      we get   ∫            1      +              1                  x          2                            1      +                        (          x          −                      1            x                    )                2              d  x  =  ∫                              (          x          −                      1            x                    )                ′            d      x              1      +                        (          x          −                      1            x                    )                2              =  arctan  ⁡      (    x    −          1      x        )    +  Cby the almost automatic integral   ∫                    f        ′            (      x      )              1      +      f      (      x              )        2              d  x  =  arctan  ⁡  f  (  x  )  +  C

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