Integral using partial fractions HELP!

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rinswa

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What is the integral of x / (4 - x2)2 dx? Using partial fraction Decomposition?

I tried factoring out the denominator like-so => -(x-2)(x+2)(4 - x2) but that didn't help me find the values of A and B.
 
Last edited:
rinswa said:
What is the integral of x / (4 - x2)2 dx? Using partial fraction Decomposition?

I tried factoring out the denominator like-so => -(x-2)(x+2)(4 - x2) but that didn't help me find the values of A and B.
Click to expand...
Out of curiousity why did you not factor the the 2nd factor of (4 - x2).

I would try letting u = (4 - x2) and getting something like int(du/u^2)
 
Jomo said:
Out of curiousity why did you not factor the the 2nd factor of (4 - x2).

I would try letting u = (4 - x2) and getting something like int(du/u^2)
Click to expand...


Well I thought if the denominator would be => -(2-x)2 (2+x)2

Then the partial fraction would turn out => x = A(-(2-x)) + B + C(2+x) + D

That just doesn't seem right & I can't do it by u-substitution since this problem requires me to perform partial fractions. Unless you can do a u-sub 1st then do partial fractions? But I haven't seen that before.
 
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rinswa said:
Well I thought if the denominator would be => -(2-x)2 (2+x)2

Then the partial fraction would turn out => x = A(-(2-x)) + B + C(2+x) + D

That just doesn't seem right.
Click to expand...
1st 4-x^2= (x-2)(x+2), so (4-x^2)^2 = (x-2)^2(x+2)^2.

Let u = 4-x^2 so du=-2xdx or -du/2 =xdx. (4-x^2)^2 = u^2

So int (xdx/(4-x^2)^2) = int (-du/2)/u^2 =(-1/2)int (du/u^2) = (-1/2) int (u^-2)du ....
 
Jomo said:
1st 4-x^2= (x-2)(x+2), so (4-x^2)^2 = (x-2)^2(x+2)^2.

Let u = 4-x^2 so du=-2xdx or -du/2 =xdx. (4-x^2)^2 = u^2

So int (xdx/(4-x^2)^2) = int (-du/2)/u^2 =(-1/2)int (du/u^2) = (-1/2) int (u^-2)du ....
Click to expand...


yes, and the answer would be something like => -1/2(1 / 4 - x2) + C or (- 1 / 2x2 + 8) + C

but this is a u-sub, I need to do this problem using the method -- partial fraction decomposition.
 
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