Hi, can someone show me how to do this using partial fractions, very appreciated. I'm having trouble with this one. Thanks.
The integral of x / 16x^4 - 1
Note that $\displaystyle \frac x{16x^4 - 1} = \frac x{(2x)^4 - 1} = \frac x{((2x)^2 + 1)((2x)^2 - 1)} = \frac x{(4x^2 + 1)(2x + 1)(2x - 1)}$
thus, the partial fractions are of the form:
$\displaystyle \frac x{16x^4 - 1} = \frac {Ax + B}{4x^2 + 1} + \frac C{2x + 1} + \frac D{2x - 1}$
can you continue?
Haha, ok, new plan. i was hoping that you could continue so that i wouldn't have to go through the horrendous calculations but now that you are forcing me to, lets see if we can make them simpler
start with a substitution:
Note that $\displaystyle \int \frac x{16x^4 - 1}~dx = \int \frac x{(4x^2)^2 - 1}~dx$
Now, let $\displaystyle u = 4x^2$
$\displaystyle \Rightarrow du = 8x~dx$
$\displaystyle \Rightarrow \frac 18~du = x~dx$
Thus, our integral becomes:
$\displaystyle \frac 18 \int \frac 1{u^2 - 1}~du$
Now it is easier to do partial fractions on this guy
Let $\displaystyle \frac 1{u^2 - 1} = \frac 1{(u + 1)(u - 1)} = \frac A{u + 1} + \frac B{u - 1}$
i think you can do that one, right? (don't forget the 1/8 ...and to back-substitute when done. we want our answer in terms of x not u)
Let me do this, for practice.
We simplify (i) by eliminating the fractions.
Multiply both sides by (16x^4 -1),
x = (Ax +B)(4x^2 -1) +C(4x^2 +1)(2x -1) +D(4x^2 +1)(2x +1) -----(ii)
Here is one way of doing this without going through all those many simultaneous equations.
(ii) being an equation, it follows that any value of x should satisfy (ii). So the idea now is to try values of x that would eliminate some of the terms in (ii). Try values of x that would result in a zero factor for some of the terms in (ii).
Examining (ii), if x = 1/2, then (4x^2 -1) = 0, and (2x -1) = 0, so
Let x = 1/2 in (ii),
1/2 = (A/2 +B)(1-1) +C(1+1)(1-1) +D(1+1)(1+1)
1/2 = 4D
D = 1/8 -------**
Same reasoning, let x = -1/2 , and substitute D = 1/8, in (ii),
-1/2 = (-A/2 +B)(0) +C(2)(-2) +(1/8)(2)(0)
-1/2 = -4C
C = 1/8 --------**
Let x = 0, and substitute C = D = 1/8, in (ii),
0 = (B)(-1) +(1/8)(1)(-1) +(1/8)(1)(1)
0 = -B
B = 0 ---------**
Let x = 1, and substitute B=0, C=D=1/8, in (ii),
1 = (A)(3) +(1/8)(5)(1) +(1/8)(5)(3)
1 = 3A +5/8 +15/8
Eliminate the fractions, multiply both sides by 8,
8 = 24A + 5 +15
A = -12/24 = -1/2 ------**
We got them all, hence, in (1),
x /(16x^4 -1) = (-x/2)/(4x^2 +1) +(1/8)/(2x +1) +(1/8)/(2x -1)
Or,
x/(16x^4 -1) = -x/[2(4x^2 +1)] +1/[8(2x +1)] +1/[8(2x -1)]
That's the decomposition.