Help with Calculate integralFind ∫ 0 a 3 x 2 − a x ( x − 2 a ) ( x 2 + a 2 ) d x I tried using partial fractions and the substitution u = a − x but I haven't made any real progress. Please help.

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Help with Calculate integralFind       ∫    0    a                      3                  x          2                −        a        x                    (        x        −        2        a        )        (                  x          2                +                  a          2                )              d  x I tried using partial fractions and the substitution   u  =  a  −  x but I haven&#039;t made any real progress. Please help.

mar1nerne · Accepted Answer

Use partial fraction            3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )        =      A          (      x      −      2      a      )        +            B      x      +      C              (              x        2            +              a        2            )      Now find A,B,C  A  =  2    B  =  1    C  =  a              3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )        =      2          (      x      −      2      a      )        +            x      +      a              (              x        2            +              a        2            )        =              3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )        =      2          (      x      −      2      a      )        +      x          (              x        2            +              a        2            )        +      a          (              x        2            +              a        2            )      So      ∫    0    a              3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )        d  x  =        ∫    0    a    (      2          (      x      −      2      a      )        +      x          (              x        2            +              a        2            )        +      a          (              x        2            +              a        2            )        )  d  x  =        (    2    ln    ⁡          |        x    −    2    a          |        +          1      2        ln    ⁡    (          x      2        +          a      2        )    +          tan              −        1              ⁡    (          x      a        )    )                      |              0    a    .

arridsd9 · Accepted Answer

Hint: Use the formular:            3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )        =      A          (      x      −      2      a      )        +            B      x      +      C              (              x        2            +              a        2            )      to resolve             3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )       into partial fraction by finding the values of A, B and C. Hope you can do that?Then      ∫    0    a              3              x        2            −      a      x              (      x      −      2      a      )      (              x        2            +              a        2            )        d  x  =      ∫    0    a        A          (      x      −      2      a      )        d  x  +      ∫    0    a              B      x      +      C              (              x        2            +              a        2            )        d  x  .

Help with Calculate integral Find <msubsup> ∫ 0 a </msubsup>

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