I need to compute the integral ∫ 2 x ( x 2 + x + 1 ) 2 ⋅ d x. I tried using the integration of a rational function technique, with A x + B x 2 + x + 1 + C x + D ( x 2 + x + 1 ) 2 , but this simply returned C = 2 and A , B , D = 0, so it doesn't really change anything.I also tried using a u substitution, setting u = x 2 + x + 1. This made the numerator 2 x = d u d x − 1, but I'm not really sure if I can do that/how to solve an integral with a derivative as a part of it.How would I go about solving this?

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I need to compute the integral   ∫                    2        x                    (                  x          2                +        x        +        1                  )          2                      ⋅  d  x. I tried using the integration of a rational function technique, with             A      x      +      B                      x        2            +      x      +      1        +            C      x      +      D              (              x        2            +      x      +      1              )        2            , but this simply returned   C  =  2 and   A  ,  B  ,  D  =  0, so it doesn&#039;t really change anything.I also tried using a   u substitution, setting   u  =      x    2    +  x  +  1. This made the numerator   2  x  =            d      u              d      x        −  1, but I&#039;m not really sure if I can do that/how to solve an integral with a derivative as a part of it.How would I go about solving this?

1s1zubv · Accepted Answer

2  ∫            x              (                  x          2                +        x        +        1                  )          2                      2  ∫            x              (        (        x        +                  1          2                          )          2                +                  3          4                          )          2                    Apply u-substitution:   u  =  x  +      1    2    2  ∫                    8        (        2        u        −        1        )                    (        4                  u          2                +        3                  )          2                      d  u  2  (  8  )  ∫                    2        u        −        1                    (        4                  u          2                +        3                  )          2                      d  uApply the Sum Rule  2  (  8  )  (  ∫                    2        u                    (        4                  u          2                +        3                  )          2                      d  u  −  ∫            1              (        4                  u          2                +        3                  )          2                      d    u    )Now,  ∫                    2        u                    (        4                  u          2                +        3                  )          2                      =  −            1              4        (        4                  u          2                +        3        )            Now,  ∫            1              (        4                  u          2                +        3                  )          2                      d    u    =          1              12                  3                      (    a    r    c    t    a    n    (          2              3              u    )    +          1      2        s    i    n    (    2    a    r    c    t    a    n    (          2              3              u    )    )    )    =  2  (  8  )  (  −            1              4        (        4                  u          2                +        3        )              −            1              12                  3                      (    a    r    c    t    a    n    (          2              3              u    )    +          1      2        s    i    n    (    2    a    r    c    t    a    n    (          2              3              u    )    )    )  After doing small calculations and substituting   u  =  x  +      1    2  ,  ∫            2      x                      x        2            +      x      +      1        d  x  =  16  (  −      1          4      (      4              x        2            +      4      x      +      4      )        −      1          24              3              (  2  a  r  c  t  a  n  (            2      x      +      1              3        )  +  s  i  n  (  2  a  r  c  t  a  n  (            2      x      +      1              3        )  )  )  )  +  C

civilnogwu · Accepted Answer

Apply integration by parts to  ∫      1                  x        2            +      x      +      1        d  xand extract your desired integral from the result.

I need to compute the integral ∫ <mstyle displaystyle="true" scriptlevel="0"

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