1
$\begingroup$

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.

  • 0
    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.2012-06-04

3 Answers 3

4

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.)

4

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

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

1

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, $(x^2-1)(x-1)+-1=x^3-x^2-x+1-1=x^3-x^2-x\not\equiv x^3+x.$