Integration of ∫ 1 x 4 + 1 d x

Question

Integration of    ∫      1                  x                  4                    +      1            d    x

Sophia Mcdowell · Accepted Answer

I think you can do it this way.                    ∫                  1                                    x              4                        +            1                                   d        x                            =                  1          2                ⋅        ∫                  2                      1            +                          x                              4                                                             d        x                                                         =                  1          2                ⋅        ∫                              (            1            −                          x                              2                                      )            +            (            1            +                          x                              2                                      )                                1            +                          x                              4                                                             d        x                                                         =                  1          2                ⋅        ∫                              1            −                          x              2                                            1            +                          x                              4                                                             d        x        +                  1          2                ∫                              1            +                          x                              2                                                          1            +                          x                              4                                                             d        x                                                         =                  1          2                ⋅        −        ∫                              1            −                          1                              x                2                                                                                        (                                      x            +                          1              x                                      )                              2                                      −            2                                   d        x        +        same trick

antennense · Accepted Answer

∫      1          1      +              x        4              d  x  =      1    2    ∫            1      +              x        2            +      1      −              x        2                    1      +              x        4              d  x  ∫            1      +              x        2                    1      +              x        4              d  x  =  ∫                    1                  x          2                    +      1                                (          x          −                      1            x                    )                2            +      2        d  xSet   x  −      1    x    =      2    tan  ⁡  ϕ  ∫            1      −              x        2                    1      +              x        4              d  x  =  −  ∫            1      −              1                  x          2                                              (          x          +                      1            x                    )                2            −      2        d  xSet   x  +      1    x    =      2    sec  ⁡  ψ

Integration of ∫ 1 x <mrow class="MJX

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