Prove formula for ∫ d x ( 1 + x 2 ) n

Question

Prove formula for   ∫            d      x              (      1      +              x        2                    )        n

Travis Fernandez · Accepted Answer

Use integration by parts,  I  =  ∫            d      x              (      1      +              x        2                    )        n              =  ∫      1          (      1      +              x        2                    )        n              ⋅  1     d  x  I  =      1          (      1      +              x        2                    )        n              ∫  1     d  x  −  ∫      (    (    −    n    )                  2        x                    (        1        +                  x          2                          )                      n            +            1                                ⋅    x    )    d  x  I  =      x          (      1      +              x        2                    )        n              +  2  n  ∫      (                  (        1        +                  x          2                )        −        1                    (        1        +                  x          2                          )                      n            +            1                                x    )    d  x  I  =      x          (      1      +              x        2                    )        n              +  2  n  ∫      (          1              (        1        +                  x          2                          )                      n                                −          1              (        1        +                  x          2                          )                      n            +            1                                )    d  x  I  =      x          (      1      +              x        2                    )        n              +  2  n  ∫            d      x              (      1      +              x        2                    )                  n                      −  2  n  ∫      1          (      1      +              x        2                    )                  n          +          1                      d  x  I  =      x          (      1      +              x        2                    )        n              +  2  n  I  −  2  n  ∫      1          (      1      +              x        2                    )                  n          +          1                      d  x  0  =      x          (      1      +              x        2                    )        n              +  (  2  n  −  1  )  I  −  2  n  ∫      1          (      1      +              x        2                    )                  n          +          1                      d  x  2  n  ∫      1          (      1      +              x        2                    )                  n          +          1                      d  x  =      x          (      1      +              x        2                    )        n              +  (  2  n  −  1  )  I  ∫            d      x              (      1      +              x        2                    )                  n          +          1                      =      x          2      n      (      1      +              x        2                    )        n              +            (      2      n      −      1      )              2      n        ∫            d      x              (      1      +              x        2                    )                  n                    setting n=n-1  ∫            d      x              (      1      +              x        2                    )                  n                      =      x          (      2      n      −      2      )      (      1      +              x        2                    )                  n          −          1                      +            (      2      n      −      3      )              2      n      −      2        ∫            d      x              (      1      +              x        2                    )                  n          −          1

istremage8o · Accepted Answer

Note            (              x                  (                      x            2                    +          1                      )                          n              −              1                                          )        ′    =            3      −      2      n              (              x        2            +      1              )                  n          −          1                      +            2      n      −      2              (              x        2            +      1              )                  n                    Then, integrate both sides to obtain  ∫            d      x              (      1      +              x        2                    )        n              =      1          2      n      −      2            x          (              x        2            +      1              )                  n          −          1                      +            2      n      −      3              2      n      −      2        ∫      1          (              x        2            +      1              )                  n          −          1                      d  x

Prove formula for ∫ d x </mrow> (

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