I want to evaluate the following integral: ∫ − ∞ 0 R x 2 + R 2 3 d x

Question

I want to evaluate the following integral:      ∫          −      ∞        0        R                            x          2                +                  R          2                    3            d    x

Rafael Dillon · Accepted Answer

The first approach gets       1    R        ∫          −      π              /            2        0    cos  ⁡  φ  d  φ  =      1    R  ; the second gets      1    R    [  x  (      x    2    +      R    2        )          −      1              /            2            ]          −      ∞        0    =  −      1    R        lim          x      →      −      ∞        x  (      x    2    +      R    2        )          −      1              /            2        =  −      1    R    ⋅  −  1  =      1    R    ,contra your sign error. You mistakenly rewrote   x  (      x    2    +      R    2        )          −      1              /            2      , rather than   −  (  1  +      R    2        /        x    2        )          −      1              /            2      , for   x  &amp;lt;  0. Bear in mind                     a        2            +      b        a    =            1      +      b              /                    a        2                    sgn      ⁡      a       for   a  ≠  0

I want to evaluate the following integral: <msubsup> ∫ <mrow class="MJX

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