partial fractions - long division

R

refid

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Dec 29, 2005
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I did long division on this integral:

. . .S (x^3 + 1) / (x^2 +3) dx

The long division gave me

. . .(x^3 + 1)/(x^2 + 3) = x - [(3x - 1) / (x^2 + 3)]

. . .S x dx - S (3x - 1) / (x^2 + 3) dx

I can't figure out how to intregate the second part:

. . .S (3x - 1) / (x^2 + 3) dx

Is my problem the initial long division?

P.S. I'm using "S" to stand for the integration symbol.
 
Try to do three:

\(\displaystyle \L
\int {xdx} + \int {\frac{1}{{x^2 + 3}}dx} + \int {\frac{{ - 3x}}{{x^2 + 3}}dx}\)

The second is arctangent form and the third is logarithmic form.
 
Thanks I got it ! How about this one

\(\displaystyle \L
\int {\frac{{ 1}}{{(x^2 - x - 2)}^2}dx}\)

\(\displaystyle \L
{\frac{{ Ax + B}}{{(x^2 - x - 2)}}} + {\frac{{ Cx + D}}{{(x^2 - x - 2)}^2}}\)

\(\displaystyle \L
(Ax + B) ( x^2 - x - 2) + Cx + D\)

\(\displaystyle \L
Ax^3 - Ax^2 - 2Ax + Bx^2 -Bx -2B + Cx + D\)

-------------------------------------------------------------
\(\displaystyle \L
-2B + D = 1\)

\(\displaystyle \L
D = 1+2B\)

\(\displaystyle \L
Ax^3 = 0\)

\(\displaystyle \L
-Ax^2 + Bx^2 = 0\)

\(\displaystyle \L
-2Ax - Bx + Cx = 0\)


So how is this so far?
 
Hello, refid!

Sorry, your set-up is incorrect . . .

\(\displaystyle \L\;\int \frac{dx}{(x^2\,-\,x\,-\,2)^2}\)
Click to expand...
The problem has linear factors.


Since \(\displaystyle \,x^2\,-\,x\,-\,2\;=\;(x\,+\,1)(x\,-\,2)\)

we have: \(\displaystyle \L\:\frac{1}{(x+1)^2(x-2)^2}\;=\;\frac{A}{x+1}\,+\,\frac{B}{(x+1)^2}\,+\,\frac{C}{x-2}\,+\,\frac{D}{(x-2)^2}\)
 
\(\displaystyle \frac{A}{x+1}\,+

\,\frac{B}{(x+1)^2}\,+

\,\frac{C}{x-2}\,+

\,\frac{D}{(x-2)^2}\)



\(\displaystyle {A}{(x+1)}{(x-2)^2}+

{B}{(x-2)^2}+

{C}{(x-2)}{(x+1)^2+

{D}{(x+1)^2}\)



\(\displaystyle {(Ax+A)}{(x-2)^2}+

{B}{(x-2)^2}+

{(Cx-2C)}{(x+1)^2+

{D}{(x+1)^2}\)



\(\displaystyle {(Ax+A)}{(x^2 - 4x + 4)}+

{B}{(x^2 - 4x + 4)}+

{(Cx-2C)}{(x^2 + 2x + 1)}+

{D}{(x^2 + 2x + 1)}\)



\(\displaystyle Ax^3 - 4Ax^2 + 4Ax +Ax^2 - 4Ax +4A + Bx^2 -4Bx + 4B + Cx^3 + 2Cx^2 + Cx -2Cx^2 -4Cx -2C + Dx^2 +2Dx + D\)


\(\displaystyle Ax^3 + Cx^3 =0\)

\(\displaystyle -4Ax^2 +Ax^2 + Bx^2 + 2Cx^2 -2Cx^2 + Dx^2 =0\)

\(\displaystyle 4Ax - 4Ax-4Bx+ Cx -4Cx +2Dx =0\)

\(\displaystyle 4A + 4B - 2C + D =1\)

-----------------------------------------------------

\(\displaystyle A= -C\)

\(\displaystyle -3A + B + D =0\) => \(\displaystyle 3C + B + D =0\)

\(\displaystyle -4B+ C -3C +2D =0\) => \(\displaystyle -4B -2C +2D =0\)

\(\displaystyle 4A + 4B - 2C + D =1\) => \(\displaystyle -6C + 4B + D =1\)


Sorry getting messy .. but i cant go further than this
 
This may help; I did with a computer algebra system.
\(\displaystyle \L
\frac{1}{{\left( {x^2 - x - 2} \right)^2 }} = \frac{1}{{9\left( {x + 1} \right)^2 }} + \frac{2}{{27\left( {x + 1} \right)}} + \frac{1}{{9\left( {x - 2} \right)^2 }} - \frac{2}{{27\left( {x - 2} \right)}}\)
 
pka said:
This may help; I did with a computer algebra system.
\(\displaystyle \L
\frac{1}{{\left( {x^2 - x - 2} \right)^2 }} = \frac{1}{{9\left( {x + 1} \right)^2 }} + \frac{2}{{27\left( {x + 1} \right)}} + \frac{1}{{9\left( {x - 2} \right)^2 }} - \frac{2}{{27\left( {x - 2} \right)}}\)
Click to expand...

i still do not understand where my mistake is
 
You should have:

\(\displaystyle \L\\A(x+1)(x-2)^{2}+B(x-2)^{2}+C(x-2)(x+1)^{2}+D(x+1)^{2}=1\)

\(\displaystyle \L\\Ax^{3}+Cx^{3}-3Ax^{2}+Bx^{2}+Dx^{2}-4Bx-3Cx+2Dx+4A+4B-2C+D=1\)

Which gives the matrix:

\(\displaystyle \L\\\left[\begin{array}{cccc|c}1&0&1&0&0\\-3&1&0&1&0\\0&-4&-3&2&0\\4&4&-2&1&1\end{array}\right]\)

Solve this and you should have the solutions pka posted.
 
umm... i havent learned how to slve matrix yet in my calculus course (still on partial fraction and taylors polynomials).. is there another way of doing this integral without matrix
 
I supplied the correct answer in hopes that you could use it to find you error. Frankly, I am not prepared to go through the tedium to solve this. Allow me to add that I think that ‘partial fractions’ is one of the most useless things we teach in calculus. Given the power of computer algebra systems a great many cherished topics of old have been made obsolete.
Here is an example of what can be done with a $40 windows CAS.
cas9zu.gif
 
Here may be method for you.

\(\displaystyle \L\\A(x+1)(x-2)^{2}+B(x-2)^{2}+C(x-2)(x+1)^{2}+D(x+1)^{2}=1\)

You have the system of equations, from which I built the matrix:

A+C=0
-3A+B+D=0
-4B-3C+2D=0
4A+4B-2C+D=1

Enter x=-1 in the first equation and you get:

\(\displaystyle \L\\B(-3)^{2}=B9\), so \(\displaystyle B=1/9\)

Enter x=2 in the equation:

\(\displaystyle D(3)^{2}=D9\), so \(\displaystyle D=1/9\)

Now, enter those into the system and solve for A and C.

You have:

A+C=0
-3A+1/9+1/9=0
-4/9-3C+2/9=0
-4A+4/9-2C+1/9=1

That makes it a little easier, don't it?.
 
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