Another Integration Problem

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

$\displaystyle du = 2x + 2$

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

$\displaystyle \int \frac{(x^{2} + 2x - 1)^{4}}{4} + C$ 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:

$\displaystyle \frac{(x^{2} + 2x - 1)}{1/2} * \frac{ (2x + 2)}{2}$

Re: Another Integration Problem

I would expand and integrate term by term.

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 $\displaystyle + 2$ that needs to be gotten rid of.

If the dx was $\displaystyle 4x dx$ 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 $\displaystyle + 2$ by subtraction.

$\displaystyle [(x^{2} + 2x - 1) - \frac{1}{2}] [(2x + 2) - 2]$

Re: Another Integration Problem

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

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

Re: Another Integration Problem

Quote:

Originally Posted by

**MarkFL2** $\displaystyle (x^2+2x-1)^3=x^6+6x^5+9x^4-4x^3-9x^2+6x-1$

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

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

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

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

Integrating term by term:

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

Now what?

Re: Another Integration Problem

I would choose to leave the 2 out front as a factor, but either way will work.

Re: Another Integration Problem

What's the next step? I did what you said. Is that how it's supposed to look so far?

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.

Re: Another Integration Problem

Reduced Fractions and added $\displaystyle + C$

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

But when do we use this method of integration as opposed to the other ways?

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.