# Integratin by Partial Fractions

• Sep 29th 2008, 01:51 PM
kevin100
Integratin by Partial Fractions
http://www.webassign.net/www21/symIm...c20739b016.gif

Here is what I've done so far...

$(x^3-4x^2-16)/((x^2)(x-4))=A/x+B/x^2+C/(x-4)$

so

$x^3-4x^2-16=A(x^2)(x-4)+B(x)(x-4)+C(x)(x^2)$

= $A(x^3-4x^2)+B(x^2-4x)+C(x^3)$

= $(A+C)(x^3)+(-4A+B)(x^2)+(-4B)(x)$

Now we have to find what the coefficients are equal to, so

A+C=1
-4A+B=-4
B=0

A=1
B=0
C=0

So now I should be able to take the integral of $1/x+0/x^2+0/(x-4)$

This equals $ln(x)$ on the interval [7,8], so my answer should be $ln(8)-ln(7)$.

This is apparently the wrong answer, so I must have made a mistake somewhere, but I'm not seeing it.
• Sep 29th 2008, 02:01 PM
Jhevon
Quote:

Originally Posted by kevin100
http://www.webassign.net/www21/symIm...c20739b016.gif

Here is what I've done so far...

$(x^3-4x^2-16)/((x^2)(x-4))=A/x+B/x^2+C/(x-4)$

so

$x^3-4x^2-16=A(x^2)(x-4)+B(x)(x-4)+C(x)(x^2)$

= $A(x^3-4x^2)+B(x^2-4x)+C(x^3)$

= $(A+C)(x^3)+(-4A+B)(x^2)+(-4B)(x)$

Now we have to find what the coefficients are equal to, so

A+C=1
-4A+B=-4
B=0

A=1
B=0
C=0

So now I should be able to take the integral of $1/x+0/x^2+0/(x-4)$

This equals $ln(x)$ on the interval [7,8], so my answer should be $ln(8)-ln(7)$.

This is apparently the wrong answer, so I must have made a mistake somewhere, but I'm not seeing it.

if the degree of the numerator is greater than or equal to the degree of the denominator, we must divide before doing partial fractions.

the highest power in the top is the same as that of the bottom. thus we must first divide the numerator by the denominator by long division of polynomials, or alternatively, note that

$\frac {x^3 - 4x^2 - 16}{x^3 - 4x^2} = \frac {x^3 - 4x^2}{x^3 - 4x^2} - \frac {16}{x^3 - 4x^2} = 1 - \frac {16}{x^3 - 4x^2}$

so your integral is $\int_7^8 \bigg( 1 - \frac {16}{x^3 - 4x^2} \bigg)~dx$

integrating 1 is easy, use partial fractions for the latter part
• Sep 29th 2008, 02:19 PM
kevin100
I'm still not getting it. My work doesn't seem to make much sense so I wont bother typing it out.
• Sep 29th 2008, 02:40 PM
Jhevon
Quote:

Originally Posted by kevin100
I'm still not getting it. My work doesn't seem to make much sense so I wont bother typing it out.

$\frac {16}{x^2(x - 4)} = \frac Ax + \frac B{x^2} + \frac C{x - 4}$

$\Rightarrow 16 = Ax(x - 4) + B(x - 4) + Cx^2$ ...............(1)

plugging in $x = 0$ in (1) we get $\boxed{B = -4}$

plugging in $x = 4$ in (1) we get $\boxed{C = 1}$

plugging in $x=1,~B=-4,$ and $C = 1$ in (1) we get $\boxed{A = -1}$
• Sep 29th 2008, 03:28 PM
Krizalid
Here.