Integration By Parts

**lolo57** · December 23rd 2010, 06:46 PM

How to integrate x³ * radical(2x+1)?

it looks like this: wolfram site

the problem says to do it using tabular integration by parts, which is what I did, but I ended up with this:
x^3/3*(2x+1)^(3/2) - 3x^2/15*(2x+1)^(5/2)+6x/105*(2x+1)^(7/2)+6/945*(2x+1)^(9/2)
I know that's really hard to see.. so here's the image: wolfram (the first box, where it says input)

it's so different from the actual answer in my calculator and wolfram! does anybody know how to integrate it properly?

**dwsmith** · December 23rd 2010, 06:49 PM

$\displaystyle u=x^3 \ \ du=3x^2dx \ \ \ dv=\sqrt{2x+1} \ \ v=\frac{(2x+1)^{3/2}}{3}$

$\displaystyle uv-\int vdu$

**lolo57** · December 23rd 2010, 06:59 PM

wait.. if dv=\sqrt{2x+1}, shouldnt v=(2x+1)^{3/2}/3?

**dwsmith** · December 23rd 2010, 07:00 PM

Yup.

**lolo57** · December 23rd 2010, 07:16 PM

hm.. but the answer I got after doing that was

$\displaystyle \frac{x^3}{3} (2x+1)^{3/2} - \frac{3x^2}{15} (2x+1)^{5/2} +\frac{6x}{105} (2x+1)^{7/2}-\frac{6}{945} (2x+1)^{9/2}+constant$

when according to my calculator and wolframalpha.com, it should be

$\displaystyle \frac{x^3}{9} (2x+1)^{3/2} - \frac{x^2}{21} (2x+1)^{3/2} +\frac{2}{105} (2x+1)^{3/2} - \frac{2}{315} (2x+1)^{3/2} +constant$

**dwsmith** · December 23rd 2010, 07:29 PM

$\displaystyle u=x^3 \ \ du=3x^2dx \ \ \ dv=\sqrt{2x+1} \ \ v=\frac{(2x+1)^{3/2}}{3}$

$\displaystyle uv-\int vdu$

$\displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\int x^2(2x+1)^{3/2}dx$

$\displaystyle s=x^2 \ \ ds=2xdx \ \ \ dt=(2x+1)^{3/2} \ \ t=\frac{(2x+1)^{5/2}}{5}$

$\displaystyle uv-\left[st-\int tds\right]$

$\displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\int \frac{2x(2x+1)^{5/2}}{5}dt$

$\displaystyle w=2x \ \ dw=2dx \ \ \ dz=\frac{(2x+1)^{5/2}}{5} \ \ z=\frac{(2x+1)^{7/2}}{35}$

$\displaystyle uv-st+\left[wz-\int zdw\right]$

$\displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\frac{2x(2x+1)^{7/2}}{35}-\int \frac{2(2x+1)^{7/2}}{35}dx$

Is this the expression you solved?

**lolo57** · December 23rd 2010, 07:44 PM

huh. um.. well actually I did the first part of it after your first post, but it looked like what I had initially tried and got, so I stopped.

the directions in the book actually told me to find the integral using tabular integration by parts, which is something like this:

and so I tried something similar, listing under the column "u"
$x^3$
$3x^2$
$6x$
$6$
$0$

and under the column "dV"
$\(2x+1)^{1/2}$
$\frac{(2x+1)^{3/2}}{3}$
$\frac{(2x+1)^{5/2}}{15}$
$\frac{(2x+1)^{7/2}}{105}$
$\frac{(2x+1)^{9/2}}{945}$

then I just connected all of them and got the result that I got. I'm not sure where I went wrong..

**dwsmith** · December 23rd 2010, 07:54 PM

If we continue from here

$\displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\frac{2x(2x+1)^{7/2}}{35}-\int \frac{2(2x+1)^{7/2}}{35}dx$ ,

we will obtain:

$\displaystyle \frac{x^3(2x+1)^{3/2}}{3}-\frac{x^2(2x+1)^{5/2}}{5}+\frac{2x(2x+1)^{7/2}}{35}-\frac{2(2x+1)^{9/2}}{315}+C$

$\displaystyle \frac{(2x+1)^{3/2}}{315}\left[105x^3-63x^2(2x+1)+18x(2x+1)^2-2(2x+1)^3\right]+C$

And continue simplifying.

Here is Mathematica 7's answer

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