I have to evaluate this integral: ∫ 1 − x + x 2 x 1 + x − x 2 d x

Question

I have to evaluate this integral:  ∫                    1        −        x        +                  x          2                            x                  1          +          x          −                      x            2                                d    x

esperoanow · Accepted Answer

You can split this into three integrals:      I    1    =  ∫      1          x              1        +        x        −                  x          2                      d  x      I    2    =  ∫      1          1      +      x      −              x        2              d  x      I    3    =  ∫      x          1      +      x      −              x        2              d  xFor       I    1  , substitute   t  =      1    x   and this leads to an acrosh-type integral.For       I    2  , this is an arcsin-type integral.For       I    3  , you can rewrite it as  −      1    2    ∫            −      2      x      +      1              1      +      x      −              x        2              d  x  +      1    2        I    2

Araceli Clay · Accepted Answer

∫                    1        −        x        +                  x          2                            x                  1          +          x          −                      x            2                                d    x  The rooted is a trinomial (  a      x          2        +  2  b  x  +  c), which has two real roots   α and   β, decomposing it we have:       a          x              2              +    2    b    x    +    c    =      a    (    x    −    α    )    (    x    −    β    )    =  (  x  −  β  )                    a        (        x        −        α        )                    x        −        β            . Let&#039;s say             a      (      x      −      α      )              x      −      β        =      t          2      , and we get :  x  =            a      α      −      β              t                  2                            a      −              t                  2                      d  x  =            2      a      t      (      α      −      β      )              (      a      −              t                  2                            )                  2                      d  tSubstituting these values in the integral, we have:   ∫                    1        −        x        +                  x          2                            x                  1          +          x          −                      x            2                                d    x    =  =  −  ∫                    4        (        2                  t                      4                          −                  t                      2                          +        2        )                    (                  t                      2                          +        1                  )                      2                          (                  t                      2                          (                  5                +        1        )        −                  5                +        1        )              d    t    =expanding  =  ∫                    2                  5                            (                  t                      2                          +        1                  )                      2                                d    t    ∫                    1        −                  5                                      t                      2                          +        1              d    t    ∫            4                        t                      2                          (                  5                +        1        )        −                  5                +        1        )              d    t  Calculating the individual integrals:      I          1        =      5    .  t  a      n          −      1        (  t  )  +                    5            t                      t                  2                    +      1            I          2        =  (  1  −      5    )  .  t  a      n          −      1        (  t  )      I          3        =  −  l  n  (            2      t      −              5            +      1              2      t      +              5            −      1        )Ultimately you have:   I  =      I          1        +      I          2        +      I          3        =  t  a      n          −      1        (  t  )  +                    5            t                      t                  2                    +      1        −  l  n  (            2      t      −              5            +      1              2      t      +              5            −      1        )All that remains is to replace the value of t with x:  t  =            −      2      x      −              5            +      1              2      x      −              5            −      1      getting the result.

I have to evaluate this integral: ∫ <mrow class="MJX-TeXAtom-ORD">

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