Put fraction in "arctan-friendly" formI would like to put ∫ 1 ( 2 x 2 + x + 1 ) d x into something like ∫ 1 ( u 2 + 1 ) d x. What is the quickest way to proceed? I know that previous fraction can be rewritten as 2 t 2 + t + 1 = 7 8 ( ( 4 t + 1 7 ) 2 + 1 ) , but I don't have any explaination from where this comes from.Finally, the integral yields ∫ b a 7 8 ( ( 4 t + 1 7 ) 2 + 1 ) d t = 2 7 [ arctan ⁡ ( 4 t + 1 7 ) ] b a

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Put fraction in &quot;arctan-friendly&quot; formI would like to put   ∫      1          (      2              x        2            +      x      +      1      )        d  x into something like   ∫      1          (              u        2            +      1      )        d  x. What is the quickest way to proceed? I know that previous fraction can be rewritten as   2      t    2    +  t  +  1  =      7    8        (                  (                              4            t            +            1                                7                          )            2        +    1    )  , but I don&#039;t have any explaination from where this comes from.Finally, the integral yields      ∫    b    a        7    8        (                  (                              4            t            +            1                                7                          )            2        +    1    )    d  t  =      2          7                  [      arctan      ⁡              (                              4            t            +            1                                7                          )            ]        b    a

sveiparnu · Accepted Answer

Note that  (  a  x  +  b      )    2    =      a    2        x    2    +  2  a  b  x  +      b    2  Now we want to complete the square on   2      x    2    +  x  +  1. We then have       a    2    =  2  ,  2  a  b  =  1    ⟹    4      a    2        b    2    =  1    ⟹        b    2    =      1    8  . Thus, we write  2      x    2    +  x  +  1  =      (    2          x      2        +    x    +          1      8        )    +      7    8    =            (              2            x      +              1                  2                      2                              )        2    +      7    8  Thus, letting       2    x  +      1          2              2              =            7      8        tan  ⁡  θ, our integral becomes  ∫      1                  7        8                    tan        2            ⁡      θ      +              7        8              ⋅            7        4        sec    2    ⁡  θ  d  θ

bu1tu2t1zt · Accepted Answer

Multiply numerator and denominator by   4  ⋅  2  =  8 (the   2 is the coefficient of       x    2  ) and “complete the square”:      1          2              x        2            +      x      +      1        =      8          16              x        2            +      8      x      +      8        =      8          16              x        2            +      8      x      +      1      +      7        =      8          (      4      x      +      1              )        2            +      7      Now you know that you should set   4  x  +  1  =  u      7  , so you get      8    7        1                  u        2            +      1      Moreover,   4    d  x  =      7      d  u and the integral becomes  ∫      8    7        1                  u        2            +      1                  7        4      d  u  =      2          7        ∫      1                  u        2            +      1          d  u  =      2          7        arctan  ⁡  u  +  c  =      2          7        arctan  ⁡      (                  4        x        +        1                    7              )    +  c

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