How to find the general antiderivative of f ( x ) = x ( 6 − x ) 2 ?I want to find the general antiderivative of f ( x ) = x ( 6 − x ) 2 . However, I keep getting it wrong.I am new to antiderivatives, but I think the first thing I should do is differentiate. d d x ( 6 − x ) 2 = 2 ( 6 − x ) ⋅ − 1 = − 2 ( 6 − x ) f ′ ( x ) = [ ( 6 − x ) 2 ⋅ 1 ] + [ − 2 ( 6 − x ) ⋅ x ] = ( 6 − x ) 2 − 2 x ( 6 − x )According to the antiderivative power rule, when n ≠ − 1, ∫ x n d x = x n + 1 n + 1 + C.So it seems like ∫ ( 6 − x ) 2 d x = ( 6 − x ) 3 3 + C.However, I can't find a rule that seems like it would work with − 2 x ( 6 − x ). The closest thing that I can find is the "Multiplication by Constant Rule", ∫ c f ( x ) d x = c ⋅ ∫ ( f ( x ) ) d x, but I'm not sure if I'm allowed to change 2 x ( 6 − x ) into the form 2 ( 6 x − x 2 ). ∫ 2 ( 6 x − x 2 ) = 2 ⋅ ∫ ( 6 x − x 2 ) = 2 ( 6 ∫ x − ∫ x 2 ) = 2 ( 6 x 2 2 − x 3 3 ) + CBut ( 6 − x ) 3 3 − ( 6 x 2 − 2 x 3 3 ) + C is incorrect.

Question

How to find the general antiderivative of   f  (  x  )  =  x  (  6  −  x      )    2  ?I want to find the general antiderivative of   f  (  x  )  =  x  (  6  −  x      )    2  . However, I keep getting it wrong.I am new to antiderivatives, but I think the first thing I should do is differentiate.      d          d      x        (  6  −  x      )    2    =  2  (  6  −  x  )  ⋅  −  1  =  −  2  (  6  −  x  )      f    ′    (  x  )  =  [  (  6  −  x      )    2    ⋅  1  ]  +  [  −  2  (  6  −  x  )  ⋅  x  ]  =  (  6  −  x      )    2    −  2  x  (  6  −  x  )According to the antiderivative power rule, when   n  ≠  −  1,   ∫      x    n    d  x  =            x              n        +        1                    n      +      1        +  C.So it seems like   ∫  (  6  −  x      )    2    d  x  =            (      6      −      x              )        3              3    +  C.However, I can&#039;t find a rule that seems like it would work with   −  2  x  (  6  −  x  ). The closest thing that I can find is the &quot;Multiplication by Constant Rule&quot;,   ∫  c  f  (  x  )  d  x  =  c  ⋅  ∫  (  f  (  x  )  )  d  x, but I&#039;m not sure if I&#039;m allowed to change   2  x  (  6  −  x  ) into the form   2  (  6  x  −      x    2    ).  ∫  2  (  6  x  −      x    2    )  =  2  ⋅  ∫  (  6  x  −      x    2    )  =  2  (  6  ∫  x  −  ∫      x    2    )  =  2  (            6              x        2              2    −            x      3        3    )  +  CBut             (      6      −      x              )        3              3    −  (  6      x    2    −            2              x        3              3    )  +  C is incorrect.

assumintdz · Accepted Answer

Step 1Another way to do it is to substitute   y  =  x  −  6, so  f  (  x  )  =  6      y    2    +      y    3    =      d          d      y            (    2          y      3        +          1      4              y      4        )    =      d          d      x                          y        3            (      y      +      8      )        4    =      d          d      x                  (      x      −      6              )        3            (      x      +      2      )        4    .Step 2Thus the general antiderivative is             (      x      −      6              )        3            (      x      +      2      )        4    +  C.