Where does this fraction come from in this integral?So you have the integral: ∫ 3 v 200 − 4 v d vI tried to do u-substitution at first with u = 200 − 4 v, but I could not get the correct answer which is: − 3 4 v − 150 4 l n ( 200 − 4 v ) + CIn the worked solution, they did not use a u-substitution. The first integral becomes: ∫ − 3 4 + 150 200 − 4 v d vAnd I cannot see what technique they used to get that. I worked out that if you actually add the 2 fractions you end up back at the first integral, but I do not see how they worked out that is the way it should be re-arranged. I also don't understand why my u-substitution didn't work. Should a u-substitution have worked? I'm still trying to get my head around this integrating of fractions.

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Where does this fraction come from in this integral?So you have the integral:  ∫            3      v              200      −      4      v        d  vI tried to do   u-substitution at first with   u  =  200  −  4  v, but I could not get the correct answer which is:  −      3    4    v  −      150    4    l  n  (  200  −  4  v  )  +  CIn the worked solution, they did not use a   u-substitution. The first integral becomes:  ∫  −      3    4    +      150          200      −      4      v        d  vAnd I cannot see what technique they used to get that. I worked out that if you actually add the 2 fractions you end up back at the first integral, but I do not see how they worked out that is the way it should be re-arranged. I also don&#039;t understand why my   u-substitution didn&#039;t work. Should a   u-substitution have worked? I&#039;m still trying to get my head around this integrating of fractions.

feaguelaBapzo · Accepted Answer

Using   u-substitution, let   u  =  200  −  4  v:  =  ∫            0.75      (      200      −      u      )        u        (    −          1      4              d        u    )    =  ∫      (          3      16        −          150              4        u              )        d    u  =      3    16    u  −      150    4    ln  ⁡      |    u      |    +  C  =      3    16    (  200  −  4  v  )  −      150    4    ln  ⁡      |    200  −  4  v      |    +  C  =      600    16    −      3    4    v  −      150    4    ln  ⁡      |    200  −  4  v      |    +  C  =  −      3    4    v  −      150    4    ln  ⁡      |    200  −  4  v      |    +  CIn the last step, the constant       600    16   is absorbed into   CThat is probably why you thought   u-substitution did not work.

Brenden Tran · Accepted Answer

3  ∫      x          200      −      4      x        d  x  =  3  ∫  [  −      25          2      (      x      −      50      )        −      1    4    ]  d  x  =  −      75    2    ∫      1          x      −      50        d  x  −      3    4    ∫  d  x  =  −      75    2    ln  ⁡  (  x  −  50  )  −      3    4    x  +  cThe first step is called &quot;long division&quot;. This is the exact solution. (Note : I used   x for   v as expression.)

Where does this fraction come from in this integral? So you have the integral: ∫<!-- ∫

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