Integration by parts

Integrate: 4p5ln(p)dp.
I set u as 4p^5, and dv as ln(p). As I'm integrating, I'm noticing that I'll be using integration by parts on lnp over and over again. After noticing this, I'm unsure about if I'm doing this problem right, or this problem is really that horrible.

Did I declare u and dv right, and is there no other way to get all the dv values other than integrating plnp 6 times?

Should be right, I didn't really simplify anything though.

Wow, I can't believe that I forgot about using lnp as u.
Thanks!

Bumping because I don't want to start a new thread.
Integrate:
3/((x + 1)(x^2 + 9))
What I got was:
(3/10)lnabs(x-1) - (3/20)(lnabs(x^2 + 9)) - (3/10)arctan(x/3) + C
but it's wrong, and I can't seem to pinpoint was I messed up on.

 Source

Code:

\begin{align*}
\int\frac{3}{(x+1)(x^2+9)}~\text{d}x& = \int\frac{A}{x+1}+\frac{Bx+C}{x^2+9}~\text{d}x \\
& = Ax^2+9A+Bx^2+Cx+Bx+C \\
& = x^2(1A+1B+0C) = 0 \\
& = x(0A+1B+1C) = 0 \\
& = (9A+0B+1C) = 3 \\
& = \text{rref}\begin{pmatrix}
1~1~0~0 \\
0~1~1~0 \\
9~0~1~3
\end{pmatrix} \\
& =  A = \frac{3}{10}~B =-\frac{3}{10}~C=\frac{3}{10} \\
& = \int\frac{\frac{3}{10}}{x+1}+\frac{-\frac{3}{10}x+\frac{3}{10}}{x^2+9}~\text{d}x \\
& = \frac{3}{10}\int\frac{1}{x+1}~-\frac{3}{10}\int\frac{x}{x^2+9}+\frac{3}{10}\int\frac{1}{x^2+9} \\
& = u = x^2+9 ~ \text{d}u ~= 2x \\
& = \frac{3}{10}\int\frac{1}{x+1}~-\frac{3}{20}\int\frac{du}{u}+\frac{1}{30}\int\frac{1}{\frac{x^2}{9}+1} \\
& = w = \frac{x}{3}~\text{d}w ~= \frac{\text{d}x}{3} \\
& = \frac{3}{10}\int\frac{1}{x+1}~-\frac{3}{20}\int\frac{du}{u}+\frac{1}{10}\int\frac{1}{w^2+1} \\
& = \frac{3\ln|x+1|}{10}-\frac{3\ln|x^2+9|}{20}+\frac{\tan^{-1}\left(\frac{x}{3}\right)}{10}+C
\end{align*}

Should be right.

stuck on another problem.
Integrate:
1/(x * sqrt(x + 1))

Used u sub setting:
u = sqrt(x +1)
x = u^2 - 1
dx = 2udu

I end up with: -16/(u * (u^2 - 1))
Move out -16, and split u^2 -1 into (u + 1)(u - 1)
Used partial fractions; got A = -1, B = 0.5, C = 0.5.

I get: -lnabs(u) + 1/2lnabs(u -1) + 1/2lnabs(u + 1) + C
Substitute x back in, multiply by the -16 I moved out, and I get:
16lnabs(sqrt(x + 1) - 8lnabs(sqrt(x + 1) - 1) - 8lnabs(sqrt(x+1) +1) + C
And that seems to be wrong.

Found my error, forgot to factor out u.

Bumping because I have a math midterm and arc lengths suck.Find the length of lnx from 1 to sqrt(3).

I'm getting u + 0.5ln(u + 1) - 0.5ln(u -1), where u = sqrt(1+ x^2), as the integral I'm plugging numbers into, but yahoo answer tells me it's sqrt(1 + x^2) - ln | [1 + sqrt(1 + x^2)] / x, so I dunno if I'm doing it right.

My first time working with arc lengths... which method are you using? I tried the standard method of Int{sqrt[1 + (dy/dx)^2]}, and it came out horribly. Had to use WolframAlpha to get that to work. Final output is: 2 - sqrt(2) + ln{[sqrt(2) + 1] / sqrt(3) }.

Hadriel