Need a find ∫ .5 2 l n ( 4 + x 2 ) d x I understand that to find this value I need to find what the power series of l n ( 4 + x 2 ) and I know how to start this off since ln ⁡ ( 1 − x ) = ∫ 1 1 − x which is 1 1 − x is the geometric series.

Question

Need a find       ∫          .5              2        l  n  (  4  +      x    2    )  d  x I understand that to find this value I need to find what the power series of   l  n  (  4  +      x    2    ) and I know how to start this off since   ln  ⁡  (  1  −  x  )  =  ∫      1          1      −      x       which is       1          1      −      x       is the geometric series.

zupa1z · Accepted Answer

Starting with   ln  ⁡  (  1  +  x  )  =  −      ∑          n      =      1        ∞    (  −  1      )    n              x      n        n   (for |x|&amp;lt;1), observe that  ln  ⁡  (  4  +      x    2    )  =  ln  ⁡  4  +  ln  ⁡      (    1    +                  x        2            4        )    =  2  ln  ⁡  2  +  ln  ⁡      (    1    +                  x        2            4        )    .In particular, we can apply the result above to   ln  ⁡  (  1  +  y  ), with   y  =            x      2        4   (note that   0  ≤  x  &amp;lt;  2 implies   0  ≤  y  &amp;lt;  1):  ln  ⁡  (  4  +      x    2    )  =  2  ln  ⁡  2  −      ∑          n      =      1        ∞              (      −      1              )        n                    n              2                  2          n                          x          2      n        .

Need a find int_(.5)^2 ln(4+x^2)dx

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