Integration with partial fractions help pleaseI'm trying to work in my partial fractions chapter and some were easy but for whatever reason, I'm stuck now: ∫ − 3 x 2 + 2 x + 4 d xWhat I tried: since my denominator is of higher order and a irreducible quadratic, I set it up to start like this: x − 3 = ∫ A x + B x 2 + 2 x + 4 + C x + D x 2 + 2 x + 4 d xI'm not really sure if that's right, the example in the book is a little different.Anyway, after multiplying and collecting terms, I end up with x 5 ( C ) = 0 x 4 ( 4 C + D ) = 0 x 3 ( A + 12 C + 4 D ) = 0 x 2 ( 2 A + B + 16 C + 12 D ) = 0 x ( 4 A + 2 B + 16 D + 16 ) = 1 4 B + 16 D = − 3but then when I tried to solve by substitution/addition methods, my numbers don't make sense. Am I even on the right track here? We just got started in this class and I feel like I'm behind already. Thanks for any help.

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Integration with partial fractions help pleaseI&#039;m trying to work in my partial fractions chapter and some were easy but for whatever reason, I&#039;m stuck now:  ∫            −      3              x      2      +      2      x      +      4        d  xWhat I tried: since my denominator is of higher order and a irreducible quadratic, I set it up to start like this:  x  −  3  =  ∫            A      x      +      B              x      2      +      2      x      +      4        +            C      x      +      D              x      2      +      2      x      +      4        d  xI&#039;m not really sure if that&#039;s right, the example in the book is a little different.Anyway, after multiplying and collecting terms, I end up with      x    5    (  C  )  =  0      x    4    (  4  C  +  D  )  =  0      x    3    (  A  +  12  C  +  4  D  )  =  0      x    2    (  2  A  +  B  +  16  C  +  12  D  )  =  0  x  (  4  A  +  2  B  +  16  D  +  16  )  =  1  4  B  +  16  D  =  −  3but then when I tried to solve by substitution/addition methods, my numbers don&#039;t make sense. Am I even on the right track here? We just got started in this class and I feel like I&#039;m behind already. Thanks for any help.

odczepneyv · Accepted Answer

You know that by substitution   y  =      x    2    +  1 you have:  ∫            2      x                      x        2            +      1        d  x  =  ∫      1    y    d  y  =  ln  ⁡      |    y      |    +  CAlso   ∫      1                  x        2            +      1        d  x  =  arctan  ⁡  x  +  CYou can use this two facts this way: write the dominator as  c  (  (            x      −      a        b        )    2    +  1  )for some   a  ,  b  ,  c and next use substitution   x  =  b  y  +  a in this case:      x    2    +  2  x  +  4  =  (  x  +  1      )    2    +  3  =  3  (  (            x      +      1              3            )    2    +  1  )Can you finish your task?

Integration with partial fractions help please I'm trying to work in my partial fractions chapter a

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