Math Help - Integration by parts

1. Integration by parts

no.1 on Flickr - Photo Sharing!
I can't get the answers they give. I keep getting

(1) 1/2 x^2 lnx - 5x lnx - 1/4 x^2 + 5x + c

u=lnx
du=1/x dx

dv=x-5
v=x^2/2 - 5x

lnx (x^2/2 - 5x) - int (x^2/2 -5x)(1/x) dx
lnx (x^2/2 - 5x) - int (x/2 - 5) dx
lnx (x^2/2 - 5x) - x^2/4 + 5x + C
1/2x^2 lnx - 5x lnx - 1/4 x^2 + 5x +C
What am I doing wrong?

Also this one has been giving me trouble
http://www.flickr.com/photos/47939227@N03/4528886766/
I keep getting
2/3 x (x^2+6)^3/2 - 4/5(x^2+6)^5/2 + C

(2) int (2x^3) (x^2 + 6)^1/2
u=2x^3
du=6x^2 dx

dv= (x^2 + 6)^1/2
v=(x^2 + 6)^3/2
--------------
3x
2/3 x (x^2 + 6)^3/2 - int ((x^2 + 6)^3/2 (6x^2)) /(3x)

2/3 x (x^2 + 6)^3/2 - int (x^2 + 6)^3/2 (2x)
2/3 x (x^2 + 6)^3/2 - 4/5 x (x^2 + 6)^5/2 + C
What am I doing wrong with this one?

Thank you very much for any help.

2. Originally Posted by bananasinpajamas
no.1 on Flickr - Photo Sharing!
I can't get the answers they give. I keep getting

(1) 1/2 x^2 lnx - 5x lnx - 1/4 x^2 + 5x + c

u=lnx
du=1/x dx

dv=x-5
v=x^2/2 - 5x

lnx (x^2/2 - 5x) - int (x^2/2 -5x)(1/x) dx
lnx (x^2/2 - 5x) - int (x/2 - 5) dx
lnx (x^2/2 - 5x) - x^2/4 + 5x + C
1/2x^2 lnx - 5x lnx - 1/4 x^2 + 5x +C
What am I doing wrong?
Nothing. That's the answer. I think wherever you took the question from has got it wrong.

Maple and Wolfram both give 1/2x^2 lnx - 5x lnx - 1/4 x^2 + 5x as the answer.

3. Also this one has been giving me trouble
no.3 on Flickr - Photo Sharing!
I keep getting
2/3 x (x^2+6)^3/2 - 4/5(x^2+6)^5/2 + C

(2) int (2x^3) (x^2 + 6)^1/2
u=2x^3
du=6x^2 dx

dv= (x^2 + 6)^1/2
v=(x^2 + 6)^3/2
--------------
3x
2/3 x (x^2 + 6)^3/2 - int ((x^2 + 6)^3/2 (6x^2)) /(3x)

2/3 x (x^2 + 6)^3/2 - int (x^2 + 6)^3/2 (2x)
2/3 x (x^2 + 6)^3/2 - 4/5 x (x^2 + 6)^5/2 + C
What am I doing wrong with this one?

Thank you very much for any help.[/QUOTE]

for No.2

Avoiding integration by parts:

$2 \int {x^3} {\sqrt(x^2+6)} dx$

Let $u = x^2 \implies du = 2 x dx$:

$= \int u {\sqrt{u+6}} du$

Again, let $v = u+6 \implies dv = du$

$= \int (v-6) \sqrt v dv$

$= \int (v^{\frac{3}{2}}-6 \sqrt v ) dv$

now integrate and then substitute the values of v and u