How to integrate ∫ d x x x 4 − 1

Question

How to integrate   ∫                    d        x                    x                              x            4                    −          1

Ryan Newman · Accepted Answer

So if you write      tan          −      1        ⁡            x      4        −    1    =  θ  ,Then  tan  ⁡  θ  =            x      4        −    1    .A right triangle that tells this story has   θ as one angle,             x      4        −    1   as the opposite side and 1 as the adjacent side. Using Pythagorean theorem we can work out the length c of the hypotenuse:  (            x      4        −    1        )    2    +      1    2    =      c    2  which shows that   c  =      x    2  So   sec  ⁡  θ  =      x    2        /    1, that is       sec          −      1        ⁡  (      x    2    )  =  θ  =      tan          −      1        ⁡  (            x      4        −    1    )  .

Zion Wheeler · Accepted Answer

We have,  ∫                    d        x                    x                              x            4                    −          1                    Let       x    2    =  sec  ⁡  (  θ  )    ⟹    2  x    d  x  =  sec  ⁡  (  θ  )  tan  ⁡  (  θ  )    d  θWhich further implies that,   d  x  =                              sec          ⁡          (          θ          )                tan        ⁡        (        θ        )            2          d  θAfter substitution, the given integral changes to:            1      2        ∫                              sec          ⁡          (          θ          )                tan        ⁡        (        θ        )                              sec          ⁡          (          θ          )                ⋅                              sec            2                    ⁡          (          θ          )          −          1                        d  θ  =            1      2        ∫                              sec          ⁡          (          θ          )                tan        ⁡        (        θ        )                              sec          ⁡          (          θ          )                ⋅                              tan            2                    ⁡          (          θ          )                        d  θ  =            1      2        ∫                              sec          ⁡          (          θ          )                tan        ⁡        (        θ        )                              sec          ⁡          (          θ          )                tan        ⁡        (        θ        )                d  θ  =            1      2        ∫  d  θ  =            1      2        θ  +  C  =                                                        1              2                                            sec                          −              1                                ⁡          (                      x            2                    )          +          C

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