collegegal
New member
- Joined
- Aug 27, 2008
- Messages
- 1
I need help solving the following integration by parts. Must use formula: f(x)*g(x) - ? f’(x)*g(x)
? x^3(lnx)dx
Thanks!!!
? x^3(lnx)dx
Thanks!!!
Using Integration by Parts for int [ x^3 (ln(x)) ] dx
- Thread starter collegegal
- Start date
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
Re: Integration by Parts
Hi collegegal:
We want to start with the complete formula for integration by parts.
? f(x) g'(x) dx = f(x) g(x) - ? g(x) f'(x) dx
We need to substitute the functions x^3, ln(x), and their first derivatives into this formula in place of the f and g notation, but what gets substituted for what?
Well, there are two choices. Both choices are valid, but one is easy and one is not.
Look at the left side of the equation: ? f(x) g'(x) dx
One choice is to substitute x^3 for f(x); we would then need to substitute ln(x) for g'(x) dx.
The other choice is to substitute ln(x) for f(x); we would then need to substitute x^3 for g'(x) dx.
Deciding which choice to make requires thinking about the other end of the equation (specifically, the part in blue below).
? f(x) g'(x) dx = f(x) g(x) - ? g(x) f'(x) dx
We usually try to choose so that the first derivative of f(x) becomes "simpler", as long as we can readily integrate g'(x) dx to get g(x).
So, for this problem, substituting ln(x) for f(x) instead of g(x) is the better choice.
f(x) = ln(x)
g'(x) dx = x^3
We can readily integrate x^3 to get (x^4)/4 for g(x).
Likewise, the first derivative of ln(x) with respect to x becomes "simpler": f '(x) dx = 1/x
Making all of these substitutions for f(x), g'(x) dx, g(x), and f'(x) dx gives the following.
? ln(x) x^3 = ln(x) (x^4)/4 - ? (x^4)/4 * (1/x)
This simplifies.
? ln(x) x^3 = (1/4)x^4 ln(x) - ? (1/4)x^3
Let us know if you need help determining ? (1/4)x^3 dx.
If we had chosen to begin with f(x) = x^3 instead, then we would have been facing the following issues.
Since g'(x) dx would be ln(x), we would need to know that g(x) is x ln(x) - x, and all of the substitutions yield this:
? x^3 ln(x) = x^3 [x ln(x) - x] - ? [x ln(x) - x] 3x^2
This choice would not be easy.
Cheers,
~ Mark
My edits: near total rewrite ...
Hi collegegal:
We want to start with the complete formula for integration by parts.
? f(x) g'(x) dx = f(x) g(x) - ? g(x) f'(x) dx
We need to substitute the functions x^3, ln(x), and their first derivatives into this formula in place of the f and g notation, but what gets substituted for what?
Well, there are two choices. Both choices are valid, but one is easy and one is not.
Look at the left side of the equation: ? f(x) g'(x) dx
One choice is to substitute x^3 for f(x); we would then need to substitute ln(x) for g'(x) dx.
The other choice is to substitute ln(x) for f(x); we would then need to substitute x^3 for g'(x) dx.
Deciding which choice to make requires thinking about the other end of the equation (specifically, the part in blue below).
? f(x) g'(x) dx = f(x) g(x) - ? g(x) f'(x) dx
We usually try to choose so that the first derivative of f(x) becomes "simpler", as long as we can readily integrate g'(x) dx to get g(x).
So, for this problem, substituting ln(x) for f(x) instead of g(x) is the better choice.
f(x) = ln(x)
g'(x) dx = x^3
We can readily integrate x^3 to get (x^4)/4 for g(x).
Likewise, the first derivative of ln(x) with respect to x becomes "simpler": f '(x) dx = 1/x
Making all of these substitutions for f(x), g'(x) dx, g(x), and f'(x) dx gives the following.
? ln(x) x^3 = ln(x) (x^4)/4 - ? (x^4)/4 * (1/x)
This simplifies.
? ln(x) x^3 = (1/4)x^4 ln(x) - ? (1/4)x^3
Let us know if you need help determining ? (1/4)x^3 dx.
If we had chosen to begin with f(x) = x^3 instead, then we would have been facing the following issues.
Since g'(x) dx would be ln(x), we would need to know that g(x) is x ln(x) - x, and all of the substitutions yield this:
? x^3 ln(x) = x^3 [x ln(x) - x] - ? [x ln(x) - x] 3x^2
This choice would not be easy.
Cheers,
~ Mark
My edits: near total rewrite ...
- Joined
- Apr 12, 2005
- Messages
- 11,339
Re: Integration by Parts
Sometimes, you can just write them down. ln(x) is a prime candidate for you f(x), since its first derivative turns it into an algebraic expression.
\(\displaystyle \int x^{3}ln(x)\;dx\;=\;\int ln(x)\;d\left(\frac{1}{4}x^{4}\right)\;=\;\frac{1}{4}x^{4}ln(x)\;-\;\int \frac{1}{4}x^{4}\;d(ln(x))\;=\;\frac{1}{4}x^{4}ln(x)\;-\;\int \frac{1}{4}x^{4}\frac{1}{x}\;dx\)
collegegal said:
Sometimes, you can just write them down. ln(x) is a prime candidate for you f(x), since its first derivative turns it into an algebraic expression.
\(\displaystyle \int x^{3}ln(x)\;dx\;=\;\int ln(x)\;d\left(\frac{1}{4}x^{4}\right)\;=\;\frac{1}{4}x^{4}ln(x)\;-\;\int \frac{1}{4}x^{4}\;d(ln(x))\;=\;\frac{1}{4}x^{4}ln(x)\;-\;\int \frac{1}{4}x^{4}\frac{1}{x}\;dx\)