Evaluate the integral ∫ 1 x 3 − 1 d x

Question

Evaluate the integral  ∫      1                  x        3            −      1        d  x

Monica Dennis · Accepted Answer

Evaluate the integral  ∫      1                  x        3            −      1        d  x  =  ∫      1          (      x      −      1      )      (              x        2            +      x      +      1      )        d  x    =  ∫      A          x      −      1        d  x  +  ∫            B      x      +      C                      x        2            +      x      +      1        d  xSolving   1  =  A  ⋅  (      x    2    +  x  +  1  )  +  (  B  x  +  C  )  (  x  −  1  )    1  =  A      x    2    +  A  x  +  A  +  B      x    2    −  B  x  +  C  x  −  C    1  =  (  A  +  B  )      x    2    +  (  A  −  B  +  C  )  x  +  A  −  C    0  =  A  +  B  ⇒  B  =  −  A    0  =  A  −  B  +  C    1  =  A  −  C  ⇒  C  =  A  −  1  0  =  A  −  (  −  A  )  +  A  −  1  ⇒  0  =  3  A  −  1  ⇒  A  =      1    3      B  =  −      1    3    ,           C  =      1    3    −  1  =  −      2    3      =      1    3    ∫            d      x              x      −      1        +  ∫            −              1        3            x      −              2        3                            x        2            +      x      +      1        d  x    =      1    3    ∫            d      x              x      −      1        −      1    3    ∫            x      +      2                      x        2            +      x      +      1        d  x    =      1    3    ln  ⁡      |    x  −  1      |    −      1    3    ∫            x      +      2              (      x      +              1        2                    )        2            +              3        4              d  x    u  =  x  +      1    2      d  u  =  d  x    =      1    3    ln  ⁡      |    x  −  1      |    −      1    3    ∫            u      −              1        2            +      2                      u        2            +              3        4              d  u    =      1    3    ln  ⁡      |    x  −  1      |    −      1    3    ∫            u      +              3        2                            u        2            +              3        4              d  u    =      1    3    ln  ⁡      |    x  −  1      |    −      1    3    ∫      u                  u        2            +              3        4              −      1    3    ∫            3      2                      u        2            +              3        4              d  u  =      1    3    ln  ⁡      |    x  −  1      |    −      1    6    ln  ⁡      |    w      |    −      1    2    ⋅  (      1                  3            2        a  r  c  t  a  g  (      2          3        u  )  )  +  c    =      1    3    ln  ⁡      |    x  −  1      |    −      1    6    ln  ⁡      |    (  x  +      1    2        )    2    +      3    4        |    −            3        3    a  r  c  t  a  g  (            2      x      +      1              3        )  +  C

Evaluate the integral int (1)/(x^3-1)dx

Answered question

Answer & Explanation

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