∫(x^4)(lnx)dx
u = ln x
du = (1/x)dx
dv = x^4 dx
v = (x^5)/5
=>
∫(x^4)(lnx)dx = (x^5)*(ln x)/5 - ∫[(x^5)/5]*(1/x)dx
∫(x^4)(lnx)dx = (x^5)*(ln x)/5 - ∫[(x^4)/5]dx
∫(x^4)(lnx)dx = (x^5)*(ln x)/5 - [(x^5)/25] + c
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2006-12-03 07:12:48
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answer #1
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answered by Luiz S 7
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To do this question, you have to use integration by parts.
Let u = ln(x)
dv = x^4 dx
du = 1/x dx
v = (1/5)(x^5)
The formula for integration by parts is
uv - Integral (vdu)
So we just replace accordingly.
ln(x)(1/5)(x^5) - Integral ( (1/5) x^5 (1/x) )dx
Note that x^5 times 1/x yields x^4
(1/5) ln(x) x^5 - Integral ((1/5)x^4)dx
Whenever we deal with integrals, we want to pull out all constants out of them because they get in the way and it's a valid step.
(1/5) ln(x) x^5 - 1/5 * Integral (x^4)dx
(1/5) ln(x) x^5 - 1/5 * [x^5/5] + C
(1/5) ln(x) * x^5 - (1/25) * x^5 + C
2006-12-03 07:11:44
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answer #2
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answered by Puggy 7
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integral {x^4 ln x } dx = x^5 [ (1/5) ln x - 1/25 ]
Luiz showed the right method for solution by parts. Checking...
d/dx { x^5 [ (1/5) ln x - 1/25 ] }
= 5 x^4 [ (1/5) ln x - 1/25 ] + x^5 [ 1/(5x) ]
= x^4 ln x - (1/5) x^4 + (1/5) x^4
= x^4 ln x
2006-12-03 07:20:30
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answer #3
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answered by Anonymous
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Well, its easy to check an answer by differentiating:
Differentiating that gives:
(1/x) * (1/5)x^5 + (ln x)(x^4) - (1/5)x^4
= (1/5)x^4 + (ln x)(x^4) - (1/5)x^4
= (ln x)(x^4).
So yes, thats what you get.
If you want the method, you'd use integration by parts. But since you've got the answer, I'm sure you've already done that..
2006-12-03 07:12:49
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answer #4
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answered by stephen m 4
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u = lnx => du/dx = 1/x
dv/dx = x^4 => v = x^5/5
Integral of (x^4)(lnx)
= uv - Integral of v du/dx dx
= (lnx)(x^5/5) - Integral of (x^5/5)(1/x) dx
= x^5lnx/5 - Integral of x^4/5 dx
= x^5lnx/5 - x^5/5(5)
= x^5lnx/5 - x^5/25
= (x^5/5)(lnx - (1/5))
2006-12-03 10:34:27
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answer #5
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answered by Kemmy 6
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int{x^4*lnx}dx
put u=lnx and dv/dx=x^4
v=int(x^4)dx=x^5/5
and du/dx=1/x
int(x^4*lnx)dx
=lnx*x^5/5-int(x^5/5*dx/x)
=x^5/5*lnx-int(x^4/5)dx
=x^5/5*lnx-x^5/25+C
=x^5/25{5lnx-1}+C
where C is a constant
i hope that this helps
2006-12-03 07:37:19
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answer #6
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answered by Anonymous
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yes, thats the correct answer
im a highschool math teacher
2006-12-03 07:12:26
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answer #7
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answered by Anonymous
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2016-11-30 02:18:39
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answer #8
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answered by ? 4
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