Help im stuck with this

$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx $

$\displaystyle \int \frac{3x^5}{x^3+1} - \frac{2x^3}{x^3+1}+ \frac{5x^2}{x^3+1}-\frac{2}{x^3+1} dx $

?

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- Jun 22nd 2007, 11:07 PM^_^Engineer_Adam^_^natural logarithm integrals
Help im stuck with this

$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx $

$\displaystyle \int \frac{3x^5}{x^3+1} - \frac{2x^3}{x^3+1}+ \frac{5x^2}{x^3+1}-\frac{2}{x^3+1} dx $

? - Jun 23rd 2007, 04:45 AMearboth
Hello,

do long division first:Code:`2x^2`

(3x^3 - 2x^3 + 5x^2 - 2) ÷ (x^3+1= 3x^2 - 2 + -------

-(3x^5 + 3x^2) x^3+1

----------------------

-2x^3 + 2x^2

-(-2x^3 -2)

------------------

2x^2

$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx$ - Jun 23rd 2007, 05:11 AM^_^Engineer_Adam^_^
but quickmath has different answer

QuickMath Automatic Math Solutions

anyway the only problem i have here is the long division...

how did it get through quickmath answer? - Jun 23rd 2007, 05:34 AMearboth
- Jun 23rd 2007, 05:45 AMearboth
Hello,

I've tried to do long division with quickmath but didn't get a reasonable result. (I assume that I don't know the appropriate command).

If you are allowed to use a TI89 then you can do long divisions by propFrac(...). Instead of the three dots write the fraction. You find this command in the F2 Algebra menu. - Jun 23rd 2007, 06:33 AMtopsquark
It

*isn't*the same! Earboth's solution didn't take into account that $\displaystyle x^3 + 1 = (x + 1)(x^2 - x + 1)$. As it happens:

$\displaystyle \frac{3x^2}{x^3 + 1} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1}$

$\displaystyle = \frac{Ax^2 - Ax + A + Bx^2 + (B + C)x + C}{x^3 + 1}$

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

Thus

$\displaystyle A + B = 3$

$\displaystyle -A + B + C = 0$

$\displaystyle A + C = 0$

Gives:

$\displaystyle A = 1$

$\displaystyle B = 2$

$\displaystyle C = -1$

So

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

Which is not the same as QuickMath's answer. But the final partial fraction split should be done. So

$\displaystyle \frac{3x^5 - 2x^3 + 5x^2 - 2}{x^3 + 1} = 3x^2 - 2 + \frac{1}{x + 1} + \frac{2x - 1}{x^2 - x + 1}$

-Dan - Jun 23rd 2007, 07:58 AMCaptainBlack
You have a transcription error at one point you have:

$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{2x^2}{x^3+1} \right)dx$

then later you have:

$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx$

The first is correct, but the latter is not.

Earboth's original decomposition is correct, and there is no need to use

partial fractions after this as we now have the sum of a polynomial and

a rational function who's numerator is a numeric multiple of the derivative

of the denominator, and so its integral can be written down immediately.

RonL - Jun 23rd 2007, 08:35 AMKrizalid
$\displaystyle \begin{aligned}\int\frac{3x^5-2x^3+5x^2-2}{x^3+1}~dx&=3\int\frac{x^2(x^3+1)}{x^3+1}~dx-2\int{dx}+\int\frac{2x^2}{x^3+1}~dx\\&=3\int{x^2}~ dx-2x+\frac23\int\frac{(x^3+1)'}{x^3+1}~dx\\&=x^3-2x+\frac23\ln\left|x^3+1\right|+k,~~k\in\mathbb{R} ~\blacksquare\end{aligned}$

- Jun 23rd 2007, 05:06 PM^_^Engineer_Adam^_^
thank you all