Integration?

Question

I need to integrate

lnx
-----dx
x^4

On the interval from 1 to infinity.

How should I begin? Should I substitute u=lnx?

Thanks!

Answer 1

ok, I have it know :

integrate( lnx / x^4 )dx

let : x = e^t >>>> dx = e^t

integrate( t e^t / e^4t) = integrate( t*e^-3t)dt

we have know : integrate( t*e^-3t)dt

Now by parts :

t = y >>>> dt = dy

e^-3t = dv >>> -1/3e^-3t = v

by parts : y*v - integrate(vdy)

-1/3t*e^-3t - integrate(-1/3*e^-3t)dt

-1/3t*e^-3t - 1/9*e^-3t

now, we made : e^t = x >>> t = lnx

-1/3*lnx*(e^-3lnx) - 1/9*(e^-3lnx)

e^-3lnx = x^-3

-1/3*lnx*x^-3 - 1/9*x^-3

the final result will be :

-x^-3 / 3 [ lnx + 1/3 ] = R(x)

you have an improper integration from 1 to infinity :

x = 1 >>> R(x) = -1/9

Lim x >>> inifinty [ -x^-3 / 3 [ lnx + 1/3 + 1/9 ]

If you solve the limit, you will have the final result : 1 / 9

yes, this is because, you replace x = 1 in the final integration result, then you apply limits when x goes to infinity in the final integration result and it's 0, then :

0 - (-1/9 ) = 1/9

hope that might helped you

Answer 2

Yes, let's do it by parts. Let u = ln x and dv = dx/x^4 . Therefore, v = -1/(3x^3), du = dx/x. So,

Int u dv = uv - Int v du = -lnx/(3 x^3) - Int (-1/(3x^3) * 1/x dx =
-lnx/(3 x^3) + Int ((3x^4) dx = -lnx/(3 x^3) +(1/3) * (-1/3) * 1/x^3 = -lnx/(3 x^3) - 1/(9x^3). If x =1, then this expression gives -1/9. And when x -> oo, it goes to 0. therefore, the integral is 0 - (-1/9) = 1/9

Answer 3

substitute u=lnx
use distributive property or multiplicative (lnx)(1/(x^4))dx
simplify.

Answer 4

Substitute u = 1/x^4 and go. FWIW, I got
-(3*ln(x) + 1)/(9*x^3) and you evaluate it ☺

Doug