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I am trying to figure out what my book did, I can't make sense of the example.

"Since the degree of the numberator is greater than the degree of the denominator, we first perform the long division. This enables us to write

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

I am mostly concerned with the transformation of the problem by long division I think.

I attempt to do this on my own.

$(x+1)$ and $(x^3 + x)$ inside the long division bracket

I am left with $x^2 - 1$ on top and a leftover -1

This is not in their answer, I do not know how they did that.

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2  
Your question seems to have $x - 1$ but you write $x + 1$ in your attempt, which is it? Also, are you remembering that $x^3 + x = x^3 + 0x^2 + x$? –  Ben Millwood Jun 4 '12 at 23:33
1  
Jordan, try to spend more time figuring out where you went wrong. If you always ask another person, you will not develop the ability to fix the mistakes by yourself, neither you will develop the ability to solve problems on your own. I'm telling you this so that you become more independent and get better at this. –  Pedro Tamaroff Jun 4 '12 at 23:49
    
I really want to but I just do not have the time to do it, I have class tomorrow and 3 more sections to cover before class. –  user138246 Jun 4 '12 at 23:53

3 Answers 3

up vote 4 down vote accepted

You did not do the long division correctly.

                    x^2 + x + 2
                _________________________
        x - 1  |    x^3       + x
                  - x^3 + x^2
                  -----------
                        + x^2 + x
                        - x^2 + x
                          --------
                               2x 
                              -2x + 2
                              -------
                                  + 2

So the quotient is $x^2 + x + 2$, and the remainder is $2$.

You can verify this by doing the product and adding the remainder: $$(x-1)(x^2+x+2) = x^3 + x^2 + 2x - x^2 - x -2 = x^3 + x - 2$$ so $$(x-1)(x^2+x+2) + 2 = x^3 + x -2 + 2 = x^3 + x.$$

Whereas you claim a quotient of $x^2-1$ and a remainder of $-1$, which would give $$(x-1)(x^2-1) -1 = x^3 -x - x^2 + 1 -1 = x^3 - x^2 - x \neq x^3 + x.$$

(Even if you tried with $x+1$ instead fo $x-1$, your answer is still incorrect, since $$(x+1)(x^2-1)-1 = x^3 - x + x^2 -1 -1 = x^3 + x^2 - x -2 \neq x^3+x.$$ If you divide by $x+1$ correctly, you'll get a quotient of $x^2-x+2$ and a remainder of $-2$, $$(x+1)(x^2-x+2)-2 = x^3 -x^2 +2x +x^2 -x + 2 -2 = x^3 +x$$ which is the correct total.)

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You want to express $x^3+x$ as $(x-1)(\text{something})+r$.

I want to show you a way of solving this problem with a homemade technique. We see that the "something" must be a polynomial of degree $2$, or either we'll be getting $x^4$ which we don't want.

$$x^3+x=(x-1)(ax^2+bx+c)+r$$

If we multiply out we get

$$x^3+x=ax^3+bx^2+cx-ax^2-bx-c+r$$

Now, we equate the coefficients in each side:

$$1x^3=ax^3$$

$$0x^2=bx^2-ax^2$$

$$1x=cx-bx$$

$$0=r-c$$

What the above means is that two polynomials are equal iff their coefficients are equal.

From the above we get $a=1$, so

$$0x^2=bx^2-x^2$$

$$1x=cx-bx$$

$$0=r-c$$

Thus $b=1$.

$$1x=cx-1x$$

Then $c=2$, and finally

$$0=r-3$$

So$r=2$. So we get what we wanted

$$x^3+x=(x-1)(x^2+x+2)+2$$

Dividing by $x-1$ gives what your book has

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

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Your first term is indeed $x^2$. Then $$x^3+x-x^2(x-1)=x^3+x-(x^3-x^2)=x^2+x,$$ and so your next term is $x$. Then $$x^2+x-x(x-1)=x^2+x-(x^2-x)=2x,$$ so your next term is $2$. Finally, $2x-2(x-1)=2$, so your remainder is $2$. I suspect you simply made an arithmetic error. It always helps to check your answer, and indeed, $$(x^2-1)(x-1)+-1=x^3-x^2-x+1-1=x^3-x^2-x\not\equiv x^3+x.$$

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