Integrate algebraic fraction with constant on top?I understand that if you have ∫ 1 x + 1 d x you simply do ln ⁡ ( x + 1 ) + C. Now I'm slight confused because in my text book, ∫ 31 x − 4 d x evaluates to 31 ln ⁡ ( x − 4 ) but ∫ − 2 4 x − 1 d x becomes − 1 2 ln ⁡ ( 4 x − 1 ). Why is this?

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Integrate algebraic fraction with constant on top?I understand that if you have   ∫      1          x      +      1        d  x you simply do   ln  ⁡  (  x  +  1  )  +  C. Now I&#039;m slight confused because in my text book,   ∫      31          x      −      4        d  x evaluates to   31  ln  ⁡  (  x  −  4  ) but   ∫            −      2              4      x      −      1        d  x becomes   −      1    2    ln  ⁡  (  4  x  −  1  ). Why is this?

bgu999dq · Accepted Answer

It is because in             −      2              4      x      −      1      , there is a   4 in the denominator. We get  ∫            −      2              4      x      −      1        d  x  =  −  2  ∫      1          4      x      −      1        d  x  =  −  2  ∫      1    4        1    u    d  u  =  −      1    2    ln  ⁡  (  u  )  =  −      1    2    ln  ⁡  (  4  x  −  1  )  +  Cwhere   u  =  4  x  −  1

Camille Flynn · Accepted Answer

When you have             −      2              4      x      −      1       take   y  =  4  x  −  1 so you have   d  y  =  4  d  x  . now, your integral is             −      2        y              d      y        4    =  −      1    2              d      y        y    =  −      1    2    ln  ⁡  (  y  )

Integrate algebraic fraction with constant on top? I understand that if you have ∫<!-- ∫

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