Tough integral

**totalnewbie** · December 9th 2006, 07:49 AM

I encountered the following integral and I am bewildered:

$\int\frac{x^5-2x^4+3}{x^3-3x^2-10x}dx$

Seems to be real hard.

**Soroban** · December 9th 2006, 08:07 AM

Hello, totalnewbie!

$\int\frac{x^5-2x^4+3}{x^3-3x^2-10x}dx$

Note that the numerator is of a higher degree than the denominator.

$\text{Long division: }\frac{x^5-2x^4+3}{x^3-3x^2-10x}\;=\;x^2+x+13 + \underbrace{\frac{49x^2+130x + 3}{x(x-5)(x+2)}}_{\text{use Partial Fractions}}$

**totalnewbie** · December 9th 2006, 09:45 AM

How do you get that

$\text{Long division: }\frac{x^5-2x^4+3}{x^3-3x^2-10x}\;=\;x^2+x+13 + \underbrace{\frac{49x^2+130x + 3}{x(x-5)(x+2)}}_{\text{use Partial Fractions}}$

**Jameson** · December 9th 2006, 11:44 AM

I'm not skilled enough at Latex to type it up for you, but have you ever done long division with polynomials? The first three terms Soroban gave you divide evenly, then the last term is the remainder.

**totalnewbie** · December 9th 2006, 11:52 AM

Originally Posted by Jameson

I'm not skilled enough at Latex to type it up for you, but have you ever done long division with polynomials? The first three terms Soroban gave you divide evenly, then the last term is the remainder.

I have never done with polynomials. I try to probe.

**topsquark** · December 9th 2006, 11:58 AM

Originally Posted by totalnewbie

How do you get that

$\text{Long division: }\frac{x^5-2x^4+3}{x^3-3x^2-10x}\;=\;x^2+x+13 + \underbrace{\frac{49x^2+130x + 3}{x(x-5)(x+2)}}_{\text{use Partial Fractions}}$

Code:

                 __x^2 + x + 13___________________
x^3 - 3x^2 - 10x ) x^5 - 2x^4 + 0x^3 + 0x^2 + 0x + 3
                 -(x^5 - 3x^4 - 10x^3)
                          x^4 + 10x^3 + 0x^2
                        -(x^4 -  3x^3 - 10x^2)
                                13x^3 + 10x^2 + 0x
                              -(13x^3 - 39x^2 -130x)
                                        49x^2 + 130x + 3

-Dan

**totalnewbie** · December 9th 2006, 03:28 PM

I got the right answer. Thank you all.

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