# Math Help - natural logarithm integrals

1. ## natural logarithm integrals

Help im stuck with this
$\int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx$

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

Help im stuck with this
$\int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx$

$\int \frac{3x^5}{x^3+1} - \frac{2x^3}{x^3+1}+ \frac{5x^2}{x^3+1}-\frac{2}{x^3+1} dx$
?
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 $u = x^3+1 \Longrightarrow \frac{du}{dx} = 3x^2$

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

3. 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?

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

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.

QuickMath Automatic Math Solutions
anyway the only problem i have here is the long division...
how did it get through quickmath answer?
Originally Posted by earboth
Hello,

do long division first:
$\int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx$
It isn't the same! Earboth's solution didn't take into account that $x^3 + 1 = (x + 1)(x^2 - x + 1)$. As it happens:
$\frac{3x^2}{x^3 + 1} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1}$

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

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

Thus
$A + B = 3$
$-A + B + C = 0$
$A + C = 0$

Gives:
$A = 1$
$B = 2$
$C = -1$

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

Which is not the same as QuickMath's answer. But the final partial fraction split should be done. So
$\frac{3x^5 - 2x^3 + 5x^2 - 2}{x^3 + 1} = 3x^2 - 2 + \frac{1}{x + 1} + \frac{2x - 1}{x^2 - x + 1}$

-Dan

7. 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 $u = x^3+1 \Longrightarrow \frac{du}{dx} = 3x^2$

$\int \frac{3x^5-2x^3+5x^2-2}{x^3+1} dx = \int \left( 3x^2 - 2 + \frac{3x^2}{x^3+1} \right)dx$
You have a transcription error at one point you have:

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

then later you have:

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

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

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