How to evaluate the integral ∫ d x x 3 + 1 ?

Question

How to evaluate the integral       ∫                  d        x                              x          3                +        1            ?

Helen House · Accepted Answer

Partial fraction expansion of             1                        x          3                +        1            :-1 is a root of the denominator, therefore,             (      x      +      1      )       is a factor:                                          x            +            1                                              x              +              1                                                (                                                                                          x                    2                                    -                  x                  +                  1                                                                      |                                                        x                    3                                    +                  0                                      x                    2                                    +                  0                  x                  +                  1                                                                                                                                                                       -                  x          3                -                  x          2                                                                             -                  x          2                                                                                                                          +                  x          2                +        x                                                                                                                            x        +        1                                                                                                                -    x    -    1  Verify the quotient&#039;s discriminant:            b      2        -    4          (      a      )              (      c      )        =    -          1      2        -    4          (      1      )              (      1      )        =    -    3  The partial fractions are as follows because there are no real roots:            1                        x          3                +        1              =                  A        x        +        B                              x          2                -        x        +        1              +          C              x        +        1            Multiply both sides by             (              x        2            -      x      +      1      )              (      x      +      1      )      :      1    =          (      A      x      +      B      )              (      x      +      1      )        +    C          (              x        2            -      x      +      1      )      Make A and B disappear by letting x = -1:      1    =    C          (                        (          -          1          )                2            -      -      1      +      1      )            C    =          1      3            1    =          (      A      x      +      B      )              (      x      +      1      )        +          1      3              (              x        2            -      x      +      1      )      Make A disappear by letting x = 0:      1    =    B    +          1      3            B    =          2      3      Let x = 1:      1    =          (      A      +              2        3            )              (      2      )        +          1      3            1    =    2    A    +          5      3            A    =    -          1      3                  1                        x          3                +        1              =          1      3                      1                    x        +        1              -          1      3                      x        -        2                              x          2                -        x        +        1            Setting up the second term for &quot;&quot;u&quot;&quot; substitution:            1                        x          3                +        1              =          1      3                      1                    x        +        1              -          1      6                      2        x        -        4                              x          2                -        x        +        1            We want       2    x    -    1   in the numerator of the second term, therefore we much create a third term for the remaining -3:            1                        x          3                +        1              =          1      3                      1                    x        +        1              -          1      6                      2        x        -        1                              x          2                -        x        +        1              -          1      6                      -        3                              x          2                -        x        +        1            Now, the first two terms will integrate to natural logarithms and the last term will be a complete the square integral to become the inverse tangent:            1                        x          3                +        1              =          1      3                      1                    x        +        1              -          1      6                      2        x        -        1                              x          2                -        x        +        1              +          1      2                      1                              x          2                -        x        +        1            Write each term as an integral:      ∫          1                        x          3                +        1                    d      x        =          1      3        ∫                  1                    x        +        1                    d      x        -          1      6        ∫                  2        x        -        1                              x          2                -        x        +        1                    d      x        +          1      2        ∫                  1                              x          2                -        x        +        1                    d      x            ∫          1                        x          3                +        1                    d      x        =          1      3              ln              (        x        +        1        )              -          1      6              ln              (                  x          2                -        x        +        1        )              +                  3            3                      tan                  -          1                            (                              2            x            -            1                                3                          )              +    C

How to evaluate the integral int dx/(x^3+1)?

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