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do you get...

lnx(1/5x^5)-1/25x^5

2006-12-03 07:04:33 · 8 answers · asked by Rob B 1 in Science & Mathematics Mathematics

8 answers

∫(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

2006-12-03 07:12:48 · answer #1 · answered by Luiz S 7 · 0 0

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 · answer #2 · answered by Puggy 7 · 0 0

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 · answer #3 · answered by Anonymous · 0 0

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 · answer #4 · answered by stephen m 4 · 0 0

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 · answer #5 · answered by Kemmy 6 · 0 0

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 · answer #6 · answered by Anonymous · 0 0

yes, thats the correct answer

im a highschool math teacher

2006-12-03 07:12:26 · answer #7 · answered by Anonymous · 0 0

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2016-11-30 02:18:39 · answer #8 · answered by ? 4 · 0 0

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