# integrating using partial fractions

• Jun 15th 2009, 02:20 PM
superdude
integrating using partial fractions
You know how the numerator must be a lower degree than the denominator and if it isn't then you need to use long division/synthetic division? That's were I'm stuck. For example $\displaystyle \int\frac{x^3+4}{x^2+4}dx$ I tried doing long division but then I got a remainder and I'm not really sure what I'm doing.

Also it kind of confuses me when it's not just one constant on the top (e.g. A/(x-9)) how does this work? For example, if one of the factors in the numerator is $\displaystyle (x-9)^3$ then would I have $\displaystyle \frac{Ax^2+Bx+C}{(x-9)^3}$?
• Jun 15th 2009, 02:39 PM
superdude
for example, how does $\displaystyle \frac{2x^5+7}{x^2+3}=2x^3-6x+\frac{18x+7}{x^2+3}$
• Jun 15th 2009, 02:40 PM
HallsofIvy
Quote:

Originally Posted by superdude
You know how the numerator must be a lower degree than the denominator and if it isn't then you need to use long division/synthetic division? That's were I'm stuck. For example $\displaystyle \int\frac{x^3+4}{x^2+4}dx$ I tried doing long division but then I got a remainder and I'm not really sure what I'm doing.

$\displaystyle \frac{x^3+ 4}{x^2+ 4}= x- 4\frac{x+4}{x^2+ 1}$. The integral of x is $\displaystyle \frac{1}{2}x^2$ minus the integral of $\displaystyle \frac{x+4}{x^2+4}$. The last can be done by "partial fractions": write it as $\displaystyle -4\frac{x}{x^2+ 4}+ \frac{1}{x^2+4}$. The first can be integrated by the substitition $\displaystyle u= x^2+ 4$ and the second can be integrated as an arctangent.

Quote:

Also it kind of confuses me when it's not just one constant on the top (e.g. A/(x-9)) how does this work? For example, if one of the factors in the numerator is $\displaystyle (x-9)^3$ then would I have $\displaystyle \frac{Ax^2+Bx+C}{(x-9)^3}$?
Surely you mean "one of the factors in the denominator is $\displaystyle (x-9)^3$. In that case you can write $\displaystyle \frac{Ax^2+ Bx+ C}{(x-9)^3}= \frac{A}{x-9}+ \frac{B}{(x-9)^2}+ \frac{C}{(x-9)^3}$.
• Jun 15th 2009, 02:43 PM
superdude
Hold on!!!!!!!!!!!!!!!!! $\displaystyle \frac{x^3+4}{x^2+4}=x-4\frac{x+4}{x^2+1}$ is what I don't get
• Jun 15th 2009, 03:13 PM
Jhevon
Quote:

Originally Posted by superdude
Hold on!!!!!!!!!!!!!!!!! $\displaystyle \frac{x^3+4}{x^2+4}=x-4\frac{x+4}{x^2+1}$ is what I don't get

the harder, but more straight forward way is to use long division or synthetic division of polynomials. it is really annoying to teach and/or type out such a solution here, so i refer you to google.

a nicer method, but one that requires more insight, is algebraic manipulation

$\displaystyle \frac {x^3 + 4}{x^2 + 4} = \frac {x^3 + 4x - 4x + 4}{x^2 + 4} = \frac {x^3 + 4x}{x^2 + 4} + \frac {-4x + 4}{x^2 + 4} = x - 4 \left( \frac {x - 1}{x^2 + 4} \right)$
• Jun 15th 2009, 03:24 PM
superdude
this sight is very good for teaching long division with polynomials. http://www.sosmath.com/algebra/factor/fac01/fac01.html

with partial fractions it confuses me when you don't just have A on the top but Ax+B. does only

for example in this video (34 sec in) the last term should be Dx+E/(1+x^2) right? that's because the x is squared? what would happen if the entire term was squared? also in the video I don't get where the A/x comes from, wouldn't it be A/x^2 ?
• Jun 15th 2009, 03:32 PM
Jhevon
Quote:

Originally Posted by superdude
this sight is very good for teaching long division with polynomials. Polynomial Long Division

with partial fractions it confuses me when you don't just have A on the top but Ax+B. does only

the idea is the numerator must be one degree less than the denominator. that's the basic principle, it does get a tad weird with awkward (non-factorable) polynomials though

Quote:

for example in this video (34 sec in) the last term should be Dx+E/(1+x^2) right? that's because the x is squared? what would happen if the entire term was squared? also in the video I don't get where the A/x comes from, wouldn't it be A/x^2 ?
where is 1 + x^2 coming from?
• Jun 15th 2009, 03:57 PM
superdude
$\displaystyle \int\frac{1}{x^2(x+1)(1+x^2)}dx$
$\displaystyle \frac{1}{x^2(x+1)(1+x^2)}=\frac{Ax+B}{x^2}+\frac{C }{x+1}+\frac{Dx+E}{1+x^2}$

is this right?
• Jun 15th 2009, 04:18 PM
Jhevon
Quote:

Originally Posted by superdude
$\displaystyle \int\frac{1}{x^2(x+1)(1+x^2)}dx$
$\displaystyle \frac{1}{x^2(x+1)(1+x^2)}=\frac{Ax+B}{x^2}+\frac{C }{x+1}+\frac{Dx+E}{1+x^2}$

is this right?

yes
• Jun 15th 2009, 04:27 PM
superdude
how do I integrate $\displaystyle \frac{4x+4}{x^2+4}$?
• Jun 15th 2009, 04:42 PM
Jhevon
Quote:

Originally Posted by superdude
how do I integrate $\displaystyle \frac{4x+4}{x^2+4}$?

write as $\displaystyle \int \frac {4x}{x^2 + 4}~dx + \int \frac 4{x^2 + 4}~dx$

use integration by substitution for the first, the second is an arctangent integral

if this is a continuation from the first problem, watch your signs!
• Jun 15th 2009, 06:56 PM
superdude
Thanks, finally I get the answer to $\displaystyle \int\frac{x^3+4}{x^2+4} dx$
There's one question I still have: after doing long division I get $\displaystyle \int x - \frac{4x+4}{x^2+4} = \int\frac{4x}{x^2+4}dx+\int\frac{4}{x^2+4}dx != \int-(\frac{4x}{x^2+4}dx+\int\frac{4}{x^2+4}dx)$ where != denotes "not equal" relation. Could someone explain to me why the minus doesn't distrubute over to the second term from the partial fraction? like if it's minus something, and then that something becomes a partial fraction, why wouldn't the second fraction be negative?
• Jun 15th 2009, 07:24 PM
Jhevon
Quote:

Originally Posted by superdude
Thanks, finally I get the answer to $\displaystyle \int\frac{x^3+4}{x^2+4} dx$
There's one question I still have: after doing long division I get $\displaystyle \int x - \frac{4x+4}{x^2+4} = \int\frac{4x}{x^2+4}dx+\int\frac{4}{x^2+4}dx != \int-(\frac{4x}{x^2+4}dx+\int\frac{4}{x^2+4}dx)$ where != denotes "not equal" relation. Could someone explain to me why the minus doesn't distrubute over to the second term from the partial fraction? like if it's minus something, and then that something becomes a partial fraction, why wouldn't the second fraction be negative?

the minus sign does distribute. which is why i told you to watch your signs last time.

what you wrote is not correct by the way, for one, where did the lone x go?
• Jun 15th 2009, 09:48 PM
superdude
the original question is integrate $\displaystyle \frac{x^3+4}{x^2+4}$
the correct answer is definatly $\displaystyle \frac{x^2}{2}-2\ln{x^2+4}+2\arctan{\frac{x}{2}}+C$

this is what I did:
by division $\displaystyle \frac{x^3+4}{x^2+4}$ becomes $\displaystyle x-\frac{4x+4}{x^2+2}$ so that can be integrated as $\displaystyle \int \frac{4x}{x^2+4}dx + 4\int\frac{1}{x^2+4}$
For the first integral I used u substition and let $\displaystyle u=x^2+4 => 2du=4x$
$\displaystyle \int \frac{1}{u}2du = 2(\ln{u})=>2\ln{x^2+4}$
and for the second integral I used trig identities after I factored out the 4
$\displaystyle \int\frac{4}{x^2+4}dx = 4\int\frac{1}{x^2+2^2}dx = \frac{4}{2}\arctan{\frac{x}{2}}$
so then plugging everything back in I would have $\displaystyle \frac{x^2}{2}-(2\ln{x^2+4}+2\arctan{x}{2}+C)$ but then the negative makes the arctan negative and the answer is off by one sign

I found out what I was doing wrong.