# Homework Help: Integration by parts x5(lnx)2 dx

1. Mar 20, 2009

### LunarJK

1. The problem statement, all variables and given/known data

Integrate the following:
(Ill use { as the integration sign)

{x5(lnx)2 dx

2. Relevant equations
{u dv = uv - {v du

I'm really new to integration by parts, and unfortunately im having to learn it out of a book for now. I sort of get the idea, but this one just doesn't look so right when im done with it.

3. The attempt at a solution
u= (lnx)2
du = 2lnx(1/x) dx
v = x6/6
dv = x5dx

{u dv = (x6/6)((lnx)2) - { (x6/6)(2 lnx)(1/x)dx
= (x6/6)((lnx)2) - (1/3) { (x5)(lnx) dx

At this point I would doing integration by parts again for the integral: { (x5)(lnx) dx.

So:
u = lnx du = 1/x dx dv = x5dx v = x6/6
--> { (x5)(lnx) dx = (lnx)(x6/6) - { (x6/6)(1/x) dx
= (lnx)(x6/6) - (1/6) { x5 dx
= (lnx)(x6/6) - x6/12

Plugging this back into my original solution:

= (x6/6)((lnx)2) - (1/3)((lnx)(x6/6) - x6/12)
=(x6/6)((lnx)2) - (lnx x6/18) - x6/36

This could even further be simplified i suppose by taking out some common factors of x6/6, but I don't even know if all this is right. Some let me know if im on the right track, or where i went wrong??

Last edited: Mar 20, 2009
2. Mar 20, 2009

### lanedance

hi LunarJK

why not try and differentiate it and see if you get the original integrand?

Last edited: Mar 20, 2009
3. Mar 20, 2009

### LunarJK

Well i tried... and i do not get the original integrand. I need some advice as to where i went wrong.

4. Mar 20, 2009

### lanedance

your method looks good I think you may have just mixed up some +- signs and factors

i get to

$$= \frac{1}{6}(x^6(\ln{x})^2 - \frac{1}{3}( x^6\ln{x} -\frac{x^6}{6}) )+c = \frac{x^6}{6.6.3}(6.\ln{x)((3.\ln{x} - 1) + 1 ))+c$$

which agrees with the online integrator
http://integrals.wolfram.com/

5. Mar 20, 2009

### lanedance

this looks good here
i think your 12 should be 6.6 = 36
this should be -(-x^6/36) = +x^6/36 (i haven;t factored in the above error as well though...)
hopefully this gets you there