# [SOLVED] evaulating the integral- partial fractions

• Dec 6th 2009, 01:30 PM
genlovesmusic09
[SOLVED] evaulating the integral- partial fractions
$\displaystyle \int_1^e {\frac{{{x^5} + 1}} {{{x^4} + 2{x^3}}}dx}$
then I divided the two and got:
$\displaystyle \int {x - 2} dx + \int {\frac{{4{x^3}}} {{{x^4} + 2{x^3}}}dx}$
$\displaystyle \frac{{4{x^3}}} {{({x^3})(x + 2)}} = \frac{{A{x^2} + Bx + C}} {{{x^3}}} + \frac{D} {{x + 2}}$
but i am stuck after that
• Dec 6th 2009, 01:38 PM
skeeter
Quote:

Originally Posted by genlovesmusic09
$\displaystyle \int_1^e {\frac{{{x^5} + 1}} {{{x^4} + 2{x^3}}}dx}$
then I divided the two and got:
$\displaystyle \int {x - 2} dx + \int {\frac{{4{x^3}}} {{{x^4} + 2{x^3}}}dx}$
$\displaystyle \frac{{4{x^3}}} {{({x^3})(x + 2)}} = \frac{{A{x^2} + Bx + C}} {{{x^3}}} + \frac{D} {{x + 2}}$
but i am stuck after that

correction ...

$\displaystyle \frac{x^5+1}{x^4+2x^3} = x - 2 + \frac{4x^3+1}{x^4+2x^3}$

and ...

$\displaystyle \frac{4x^3+1}{x^3(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x+2}$
• Dec 6th 2009, 01:48 PM
Krizalid
$\displaystyle \frac{x^{5}+1}{x^{4}+2x^{3}}=\frac{x^{5}}{x^{3}(x+ 2)}+\frac{1}{x^{3}(x+2)}=\frac{x^{2}}{x+2}+\frac{1 }{x^{3}(x+2)}.$

so in order to integrate the first term we have $\displaystyle \frac{x^{2}}{x+2}=\frac{x^{2}-4+4}{x+2}=x-2+\frac{4}{x+2}.$

and as for solving $\displaystyle \int{\frac{dx}{x^{3}(x+2)}},$ put $\displaystyle x=\frac1t$ and the integral becomes $\displaystyle -\int{\frac{t^{2}}{1+2t}\,dt}=-\frac{1}{4}\int{\frac{1-\left( 1-4t^{2} \right)}{1+2t}\,dt}.$

now, can you finish this?

(i definitely dislike partial fractions.)
• Dec 6th 2009, 01:50 PM
Jester
Quote:

Originally Posted by Krizalid
$\displaystyle \frac{x^{5}+1}{x^{4}+2x^{3}}=\frac{x^{5}}{x^{3}(x+ 2)}+\frac{1}{x^{3}(x+2)}=\frac{x^{2}}{x+2}+\frac{1 }{x^{3}(x+2)}.$

so in order to integrate the first term we have $\displaystyle \frac{x^{2}}{x+2}=\frac{x^{2}-4+4}{x+2}=x-2+\frac{4}{x+2}.$

and as for solving $\displaystyle \int{\frac{dx}{x^{3}(x+2)}},$ put $\displaystyle x=\frac1t$ and the integral becomes $\displaystyle -\int{\frac{t^{2}}{1+2t}\,dt}=-\frac{1}{4}\int{\frac{1-\left( 1-4t^{2} \right)}{1+2t}\,dt}.$

now, can you finish this?

(i definitely dislike partial fractions.)

Yes - we know! (Rofl)
• Dec 6th 2009, 02:51 PM
genlovesmusic09
Quote:

Originally Posted by skeeter
correction ...

$\displaystyle \frac{x^5+1}{x^4+2x^3} = x - 2 + \frac{4x^3+1}{x^4+2x^3}$

and ...

$\displaystyle \frac{4x^3+1}{x^3(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x+2}$

then do I use a matrix to solve?
and if so do you know how to plug it into a ti-83?