# Thread: some help with this integration by parts problem

1. ## some help with this integration by parts problem

ok, i am pretty comfortable with both my integration by parts, and much moreso my u-substitution

today we went over integration by parts (i had already learned it in a much more challenging course, but for some reason my school is making me take the easier version too)... the problem was: integral of [x^2] [log (x-1)^.5] dx

i got an answer that made sense using 2 waves of integration by parts..it yielded a very ugly answer (i dont have my notebook handy, but i remember one of the terms had like a 115/8 coeficcient or something)...
then i redid the problem with u-substitution, and got a much simpler answer (i set u=x-1, and therefore x=u+1, and obviously du=dx)...is it possible these 2 very different answers are equivalent? if not, i have no idea where i went wrong (i must have gone wrong with at least one of the methods if this is the case)... can someone do this with both methods and see if they get the same thing- hopefully it shouldnt take too long- each way took less than 5 mins.. ill try to find my notebook and put exactly what answers i got, but i cant imagine what i did wrong to get such drastically differing answers

2. Could you be more careful with your parentheses, please? For example, are you integrating

$\displaystyle \displaystyle{\int x^{2}\ln(\sqrt{x-1})\,dx}$, or

$\displaystyle \displaystyle{\int x^{2}\sqrt{\ln(x-1)}\,dx}$?

3. $\displaystyle \int x^2 * log( [x-1]^\frac{1}{2}) dx$

$\displaystyle = \int x^2 * \frac{1}{2} * log(x-1) dx$

$\displaystyle = \frac{1}{2} \int x^2 * log(x-1) dx$

Let $\displaystyle u = log(x-1)$ and $\displaystyle dv = x^2$

$\displaystyle \frac{1}{6}( x^3 * log(x-1) - \int( \frac{x^3}{x-1} dx)$

Solving the second integral:
$\displaystyle \int( \frac{x^3}{x-1} dx )$

= $\displaystyle \int \frac{ x(x+1)(x-1)}{x-1} + \frac{x}{x-1} dx$

= $\displaystyle \int (x^2 + x)dx + \int \frac{x}{x-1}dx$

= $\displaystyle \frac{x^3}{3} + \frac{x^2}{2} + x + log(x-1)$

Plug that back into the first part and we get

$\displaystyle \frac{1}{6}(x^3 * log(x-1) - \frac{x^3}{3} - \frac{x^2}{2} - x - log(x-1)) + C$

Which can be prettied up some.

Edit: Going to clean it up to make it easier for you to check

$\displaystyle \frac{1}{6}(log(x-1)(x^3 - 1) -x( \frac{x^2}{3} + \frac{x}{2} + 1)) + C$

4. damn i forgot my notebook in my car again... ill be back hopefully early tomorrow to check my answers... i know my integration by parts was much more complicated