Where does this fraction come from in this integral?So you have the integral: ∫ 3...

Sattelhofsk

Sattelhofsk

Answered

2022-06-19

Where does this fraction come from in this integral?
So you have the integral:
3 v 200 4 v d v
I 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 ) + C
In the worked solution, they did not use a u-substitution. The first integral becomes:
3 4 + 150 200 4 v d v
And 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.

Answer & Explanation

feaguelaBapzo

feaguelaBapzo

Expert

2022-06-20Added 9 answers

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 | + C
In the last step, the constant 600 16 is absorbed into C
That is probably why you thought u-substitution did not work.
Brenden Tran

Brenden Tran

Expert

2022-06-21Added 9 answers

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 + c
The first step is called "long division". This is the exact solution. (Note : I used x for v as expression.)

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