Solving the indefinite integral ∫ 1 x 2 + x + 1 d xI want to solve the following indefinite integral: (1) ∫ 1 x 2 + x + 1 d xcompleting the square: (2) = ∫ 1 ( x + 1 2 ) 2 + 3 4 d xSubstitution: (3) u = 2 x + 1 3 brings: (4) d x = 3 2 d u(5) = 2 3 ∫ 1 u 2 + 1 d uThen getting standard integral (6) = arctan ⁡ ( u )and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don't know where the 3 beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?

Question

Solving the indefinite integral   ∫      1                  x        2            +      x      +      1        d  xI want to solve the following indefinite integral: (1)  ∫      1                  x        2            +      x      +      1        d  xcompleting the square: (2)  =  ∫      1          (      x      +              1        2                    )        2            +              3        4              d  xSubstitution: (3)  u  =            2      x      +      1              3      brings: (4)  d  x  =            3        2    d  u(5)  =      2          3        ∫      1                  u        2            +      1        d  uThen getting standard integral (6)  =  arctan  ⁡  (  u  )and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don&#039;t know where the       3   beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?

nizkem0c · Accepted Answer

(      x      +              1        2            )        2    +      3    4    =            (                        2          x          +          1                2            )        2    +            (                        3                2            )        2    =      3    4        (                  (                              2            x            +            1                                3                          )            2        +    1    )    =      3    4    (      u    2    +  1  )  .Since             2      x      +      1              3        =  u, we have            (                        2          x          +          1                          3                    )        ′    d  x  =  d  uor      2          3        d  x  =  d  uor  d  x  =            3        2    d  u  .Thus,  ∫      1                  x        2            +      x      +      1        d  x  =  ∫      1                  3        4            (              u        2            +      1      )        ⋅            3        2    d  u  =      2          3        ∫      1          1      +              u        2              d  u  .

Solving the indefinite integral int 1/(x^2+x+1) dx

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