# natural logarithm integrals

• Jun 22nd 2007, 11:07 PM
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 AM
earboth
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Help im stuck with this
$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx$

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

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
Now use integration by substitution. Let $\displaystyle u = x^3+1 \Longrightarrow \frac{du}{dx} = 3x^2$

$\displaystyle \int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx$
• Jun 23rd 2007, 05:11 AM
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 AM
earboth
Quote:

Originally Posted by ^_^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?

Hello,

the denominator you used with quickmath is not the same you published with your problem
• Jun 23rd 2007, 05:45 AM
earboth
Quote:

Originally Posted by ^_^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?

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 AM
topsquark
Quote:

Originally Posted by ^_^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?

Quote:

Originally Posted by earboth
Hello,

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

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 AM
CaptainBlack
Quote:

Originally Posted by earboth
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
Now use integration by substitution. Let $\displaystyle u = x^3+1 \Longrightarrow \frac{du}{dx} = 3x^2$

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

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 AM
Krizalid
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Help im stuck with this

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

\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