Evaluating ∫ 0 ∞ ln ⁡ x x 2 + 2 x + 2 d xI tried to use completing the square and use the answers in the linked question but since the log term in the numerator has x not x + 1(after completing the square the quadratic is ( x + 1 ) 2 + 1) so that doesn't seem to work.

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Evaluating       ∫    0    ∞              ln      ⁡      x                      x        2            +      2      x      +      2        d  xI tried to use completing the square and use the answers in the linked question but since the log term in the numerator has   x not   x  +  1(after completing the square the quadratic is   (  x  +  1      )    2    +  1) so that doesn&#039;t seem to work.

artyleriaCuy · Accepted Answer

Let   I be our integral. First, change of variable   x  =  2  t gives:  I  =      ∫    0          +      ∞                  log      ⁡      t              2              t        2            +      2      t      +      1        +  log  ⁡  2      ∫    0          +      ∞                  d      t              2              t        2            +      2      t      +      1      In the first integral, change of variable   t  =  1      /    u      ∫    0          +      ∞                  log      ⁡      t              2              t        2            +      2      t      +      1        =  −      ∫    0          +      ∞                  log      ⁡      u                      u        2            +      2      u      +      2        d  u  =  −  IHence  2  I  =  log  ⁡  2      ∫    0          +      ∞                  d      t              2              t        2            +      2      t      +      1        =  log  ⁡  (  2  )  [      A    r    c    t    a    n    (  2  t  +  1  )      ]    0          +      ∞        =      π    4    log  ⁡  2

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