# Another Integration Problem

• Nov 21st 2012, 08:09 PM
Jason76
Another Integration Problem
$\int(x^{2} + 2x - 1)^{3}2x dx$

$du = 2x + 2$

Now putting x to the 4th power, and 4 in the denominator.

$\int \frac{(x^{2} + 2x - 1)^{4}}{4} + C$ Not the Final Answer

However, that's the not the final answer. Some special manipulation needs to be done. That's because the du is larger than dx (of the original problem). It think some sort of division needs to be tried.

Let's try it this way:

$\frac{(x^{2} + 2x - 1)}{1/2} * \frac{ (2x + 2)}{2}$
• Nov 21st 2012, 08:33 PM
MarkFL
Re: Another Integration Problem
I would expand and integrate term by term.
• Nov 21st 2012, 08:36 PM
Jason76
Re: Another Integration Problem
Quote:

Originally Posted by MarkFL2
I would expand and integrate term by term.

How? Give me a head start. But don't do the whole thing.

Note: The du almost matches the dx (in the original problem). However, there is an extra $+ 2$ that needs to be gotten rid of.

If the dx was $4x dx$ then division by 2 would be the key. But in this problem, that isn't the case.

Ok, let's try to get rid of the $+ 2$ by subtraction.

$[(x^{2} + 2x - 1) - \frac{1}{2}] [(2x + 2) - 2]$
• Nov 21st 2012, 08:46 PM
MarkFL
Re: Another Integration Problem
$(x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1$

Now, distribute the $x$ onto that, pull the two out front, then integrate term by term.
• Nov 21st 2012, 08:54 PM
Jason76
Re: Another Integration Problem
Quote:

Originally Posted by MarkFL2
$(x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1$

Now, distribute the $2x$ onto that, pull the two out front, then integrate term by term.

$\int(x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1$

$\int 2x(x^6+6x^5+9x^4-4x^3-9x^2+6x-1)$

$\int 2x^{7} + 12x^{6} + 9x^{5} - 8x^{4} - 18x^{3} + 12x^{2} - 2x$

Integrating term by term:

$\int \frac{2x^{}8}{8} + \frac{12x^{7}}{7} + \frac{9x^{6}}{6} - \frac{8x^{5}}{5} - \frac{18x^{4}}{4} + \frac{12x^{3}}{3} - \frac{2x^{2}}{2}$

Now what?
• Nov 21st 2012, 09:03 PM
MarkFL
Re: Another Integration Problem
I would choose to leave the 2 out front as a factor, but either way will work.
• Nov 21st 2012, 09:05 PM
Jason76
Re: Another Integration Problem
What's the next step? I did what you said. Is that how it's supposed to look so far?
• Nov 21st 2012, 09:10 PM
MarkFL
Re: Another Integration Problem
Now, simply integrate term by term using the power rule. I see you have edited your post...once you have integrated, you don't want to use the integration symbol, that is removed when you write the anti-derivative. So, remove the integration symbol, reduce any fractions and add the constant of integration.
• Nov 21st 2012, 09:22 PM
Jason76
Re: Another Integration Problem
Reduced Fractions and added $+ C$

$\frac{x^{}8}{4} + \frac{12x^{7}}{7} + \frac{3x^{6}}{2} - \frac{8x^{5}}{5} - \frac{9x^{4}}{2} + \frac{4x^{3}}{1} - \frac{x^{2}}{1} + C$

But when do we use this method of integration as opposed to the other ways?
• Nov 21st 2012, 09:51 PM
MarkFL
Re: Another Integration Problem
Where the denominators are 1, I would omit them.

I looked at methods of substitution, but did not readily find a method that would work. There may be a way, but my inclination was to just expand and integrate.