Antiderivative fractionI'm would like to find the antiderivative of x 2 ( x − 1 ) 5 I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving x ² ( x − 1 ) 5 = 1 ( x − 1 ) 3 + 2 ( x − 1 ) 4 + 1 ( x − 1 ) 5 Could you tell me how to proceed?

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Antiderivative fractionI&#039;m would like to find the antiderivative of             x      2              (      x      −      1              )        5            I tried without using partial fractions but I did not manage, so I started a lesson on partial fractions but I do not succeed in proving            x              ²                    (      x      −      1              )        5              =      1          (      x      −      1              )        3              +      2          (      x      −      1              )        4              +      1          (      x      −      1              )        5            Could you tell me how to proceed?

iceniessyoy · Accepted Answer

Step 1This is a different way of looking at David&#039;s solution, but without doing a substitution. You want factors of   (  x  −  1      )    r   in the numerator, so you can cancel them with the denominator.Looking at the highest term, which is       x    2  , it seems that   (  x  −  1      )    2   is a good place to start, because it takes care of the highest order term.Write       x    2    =  (  x  −  1      )    2    +  p  (  x  ) then   p  (  x  )  =      x    2    −  (  x  −  1      )    2    =  2  x  −  1Step 2Now   p  (  x  )  =  2  (  x  −  1  )  +  1 so that       x    2    =  (  x  −  1      )    2    −  2  (  x  −  1  )  +  1 and you get the three terms you are looking for.This looks like a trick. The method of partial factions (which has its own tricks for efficient computation) automates such tricks in a way which always works. Occasionally there is a simpler way of looking at things, like the substitution here. Only practice will help you to develop efficient methods and notice short cuts.

Summer Bradford · Accepted Answer

Step 1Indeed you are correct that,             x      2              (      x      −      1              )        5              =      1          (      x      −      1              )        3              +      2          (      x      −      1              )        4              +      1          (      x      −      1              )        5            Step 2One can integrate each of these terms in turn. I will do the first to help. Let   u  =  x  −  1 and   d  u  =  d  x then,  ∫      1          (      x      −      1              )        3              d  x  =  ∫      u          −      3        d  u  =  −      1    2        u          −      2        =  −      1    2        1          (      x      −      1              )        2              +  c where c is the constant of integration. The final answer is,   −      1    2        1          (      x      −      1              )        2              −      2    3        1          (      x      −      1              )        3              −      1    4        1          (      x      −      1              )        4              +  c

Antiderivative fraction I'm would like to find the antiderivative of x 2

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