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Topic: Integral (Read 2219 times) |
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mathstudent155
Newbie
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Posts: 1
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Hello everybody!
May I ask you a big favor?
I need help.
What I have is:
(2x-3)/ sqrt(4x-1)
What is the integral of it??
I've tried to use Integral Calculator here:
numberempire.com/integral...var=x&answers=
And I've got : (x-4)*sqrt(4*x-1)/3
Seems like it is the right answer.
But I want to understand the way of solving it.
Please show me step by step solution.
Thanks in advance!
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: Integral
« Reply #1 on: Jan 27th, 2013, 1:08pm » |
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You can use integration by parts. http://en.wikipedia.org/wiki/Integration_by_parts
We can take (2x-3)/ sqrt(4x-1) = (2x-3) d[1/2 sqrt(4x-1)]
Then  (2x-3) d[1/2 sqrt(4x-1)]
= 1/2 sqrt(4x-1) * (2x-3) - 1/2 sqrt(4x-1) d[2x-3]
= 1/2 sqrt(4x-1) * (2x-3) - sqrt(4x-1) d x
= 1/2 sqrt(4x-1) * (2x-3) - 1/4 * 2/3 * (4x-1)^(3/2)
= ((2x-3)*3-(4x-1))/6 * sqrt(4x-1)
= (x-4)/3 * sqrt(4x-1)
(I think it's literally been years since I've last done this..)
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| « Last Edit: Jan 27th, 2013, 1:15pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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tsitut
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Posts: 5
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Re: Integral
« Reply #2 on: Feb 20th, 2013, 12:12pm » |
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Thanks, I actually needed it too 
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