# Thread: Integrate using partial fractions

1. ## Integrate using partial fractions

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

2. Originally Posted by element
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?

3. Thanks, I got to that step. But I'm having trouble solving for A,B,C and D.

4. Originally Posted by element
Thanks, I got to that step. But I'm having trouble solving for A,B,C and D.
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

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)

5. Thank you very much Jhevon! That made it a whole lot easier. Yes I can go on from here. Thank you.

6. Originally Posted by element
Thank you very much Jhevon! That made it a whole lot easier. Yes I can go on from here. Thank you.
You're very much welcome, Sir...or Ma'am...

7. Originally Posted by Jhevon

$\displaystyle \frac x{16x^4 - 1} = \frac {Ax + B}{4x^2 + 1} + \frac C{2x + 1} + \frac D{2x - 1}$ -----------(i)
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.

8. Originally Posted by ticbol
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 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.
good to see you again, ticbol! i sent you an email (you don't allow PMs), did you get it?

9. Originally Posted by ticbol
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.
Still in excellent form, I see.

Just one small thing .... About the bit I highlighted in red .....

It's not actually an equation, it's an identity (or an equivalence statement). (Equations aren't true for all values, but identities are).

10. Originally Posted by mr fantastic
Still in excellent form, I see.

Just one small thing .... About the bit I highlighted in red .....

It's not actually an equation, it's an identity (or an equivalence statement). (Equations aren't true for all values, but identities are).
Hey, mr fantastic!

Geez, I'm not yet into those exacting statements. I am just an civil engineering graduate so I do not dabble into those equation/identity/equality/whatever. Those are for pure mathematicians only.

But I'm picking up in my area of Math, eh?

11. Originally Posted by ticbol
Hey, mr fantastic!

Geez, I'm not yet into those exacting statements. I am just an civil engineering graduate so I do not dabble into those equation/identity/equality/whatever. Those are for pure mathematicians only.

But I'm picking up in my area of Math, eh?
...and i'm completely ignored again...

12. Originally Posted by Jhevon
...and i'm completely ignored again...

13. Originally Posted by Jhevon
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

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)
To the OP, it might be well to learn this little short cut

$\displaystyle \int\frac{dx}{1-x^2}={\rm{arctanh}}(x)$

14. Originally Posted by Mathstud28
To the OP, it might be well to learn this little short cut

$\displaystyle \int\frac{dx}{1-x^2}={\rm{arctanh}}(x)$
going into hyper-drive, i see. ok, i see what game you're playing

15. Originally Posted by Jhevon
going into hyper-drive, i see. ok, i see what game you're playing
Nice, haha.