I guess this question fits here, right then, I am stuck with pure mathematics, how the hell am I supposed to integrate lnx/x^2 ??? ive spent hours and hours it just doesn't integrate right!
Ive been using integration by parts, which is required by the question
INT u dv/dx dx = uv - INT v du/dx dx
Integration? - Math
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- Velociraptor
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Re: Integration? - Math
Sure, here we go:
So you have f(x) = ln(x)x^-2
Let u = ln(x)
So, du/dx = x^-1
Thus, du = x^-1 dx
Now, let dv/dx = x^-2
Therefore, integrate dv/dx to get v = -x^-1
Now for integration by parts:
INT u.dv = uv - INT v.du -Sub in the above results.
= ln(x).-x^-1 - INT -x^-1 . x^-1 . dx
= ln(x).-x^-1 - INT -x^-2 . dx
= ln(x).-x^-1 - x^-1 + C - But lets just ignore the constant C.
= ln(x).-x^-1 - x^-1
Therefore, INT u.dv = ln(x).-x^-1 - x^-1 = -x^-1(ln(x)+1)
Hope that helps.
So you have f(x) = ln(x)x^-2
Let u = ln(x)
So, du/dx = x^-1
Thus, du = x^-1 dx
Now, let dv/dx = x^-2
Therefore, integrate dv/dx to get v = -x^-1
Now for integration by parts:
INT u.dv = uv - INT v.du -Sub in the above results.
= ln(x).-x^-1 - INT -x^-1 . x^-1 . dx
= ln(x).-x^-1 - INT -x^-2 . dx
= ln(x).-x^-1 - x^-1 + C - But lets just ignore the constant C.
= ln(x).-x^-1 - x^-1
Therefore, INT u.dv = ln(x).-x^-1 - x^-1 = -x^-1(ln(x)+1)
Hope that helps.
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Re: Integration? - Math
I had thought this would be a question regarding numerical integration and computers, given the section where it was being asked... turned out to be simple mathematical integration instead.
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Re: Integration? - Math
thanks god I gave up maths after my GCSEs
post something intelligent before I go insane from a combination of exhaustion and boredom
Re: Integration? - Math
Hey thanks Chris! I had figured out how to do this, iam still learning to integrate by parts, turned out i wasn't using the formula properly. But thanks again! Iam sure of the way to use the formula now.
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Re: Integration? - Math
This makes me miss math
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Re: Integration? - Math
Everyone's favorite integral:
INT sqrt(tanx)dx
INT sqrt(tanx)dx
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