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Integration by parts |
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| Sep1-08, 11:57 AM | #1 |
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Integration by parts
1. The problem statement, all variables and given/known data
[tex]\int[/tex]ln(7x+9)dx 2. Relevant equations derivative of ln is 1/x 3. The attempt at a solution Well im just learning IBP so i set u=7x+9 dv=lndx but im stuck there. How do you know which to make ur u is there a way or is it trial and error Can i split the integral as: ∫ln(7x)+∫9 |
| Sep1-08, 12:04 PM | #2 |
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Recognitions:
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If you're doing by parts, note that the integrand expression (ln(7x+9)) must be comprised of udv, where u*dv. So u cannot be inside the ln expression. Let u=ln(7x+9). Continue from here.
In general, you should let u be an expression easy to differentiate and dv be the expression easy to integrate, and make sure that vdu is easy to integrate as well. It's some trial and error which gets better with experience. |
| Sep1-08, 12:13 PM | #3 |
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u=ln(7x+9)
du=1/(x+9)dx dv=dx v=x using the IBP formula i get: ln(7x+9)x- (1/7)∫x*(1/x+9) dx is this right or have i made a mistake |
| Sep1-08, 12:23 PM | #4 |
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Recognitions:
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Integration by parts
du is wrong.
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| Sep1-08, 12:32 PM | #5 |
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du=1/(x+9)*7 dx
ok so my answer i got is 1/7(7x+9)(log(7x+9)1) |
| Sep1-08, 12:33 PM | #6 |
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Recognitions:
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Your du is still incorrect. How do you differentiate ln(f(x))?
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| Sep1-08, 12:42 PM | #7 |
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what do u mean how should it be
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