Question

Can someone show how to finish this/ where I'm going wrong. I have the integral: dx/ (25-x^2)^2. I had the answer as [ln(5+x)-ln(5-x)] / 10 + C. The teacher said that was wrong, and to do the following: (A/5-x)+(B/5+x)+(C/(5-x)^2)+ (D/(5+x)^2). Can someone show this step and how to get the final product? Makes 0 sense without seeing it I'm stuck..

Answer 1

This is because you need to seperate all your things to partical fractions first.

Answer 2

It's a bit of a complicated example really, but i would say that 1/ (25-x^2)(25-x^2) can be split further

Answer 3

I had tried partial fractions but did it wrong twice according to prof

Answer 4

Okk, so one you have the 1/(25-x^2) * 1/(25-x^2), each one of them is split again.
it goes to 1/(5+x)(5-x) * 1/(5+x)(5-x) ok.
Then you can put it together under sort of this thing
1/(5+x)^2 * 1/(5-x)2

Answer 5

And when you have that, by partial fraction rules 1/(x+5)2 = A/(x+5) + B(x+5)^2

Answer 6

I'm not completely sure if that's right, but would that work?

Answer 7

Unless I'm missing something doesn't that just give the same final answer I already have though?

Answer 8

(A/5-x)+(B/5+x)+(C/(5-x)^2)+ (D/(5+x)^2). thats basically what you end up with. Did you calculate all the letters for these?

Answer 9

I must be doing something wrong cause i came up with virtually the same answer. Can you walk me through that?

Answer 10

1/(5+x)^2 * 1/(5-x)2
1 = A(5+x) + B(x+5)^2 + C(5-x) + D(5-x)^2
if X= -5 then
1 = 10C + 100D

did you do all the equations like these until you found all the letters?

Answer 11

not exactly no lol

Answer 12

Well your way is probably way easier, but that was how I had to learn it. I need to find all the letters A,B,C,D and they will be different, not nessesary all of them 1.

Answer 13

No worries, I'd prefer to see your way to the answer so I can learn better since I'm failing at it

Answer 14

@arilove1d

Answer 15

Once you have 1=10C+100D what did you do next equation wise?

Answer 16

I have the final answer as (1/10)*ln[(5 + x)/(5 - x)] + C after that is this correct?

Answer 17

I have to go to work but any helpful posts are appreciated!

Answer 18

\[\begin{align*}\frac{1}{(25-x^2)^2}&=\frac{A}{5-x}+\frac{B}{(5-x)^2}+\frac{C}{5+x}+\frac{D}{(5+x)^2}\\
1&=A(5-x)(5+x)^2+B(5+x)^2+C(5+x)(5-x)^2+D(5-x)^2\\
1&=A(125+25x-5x^2-x^3)+B(25+10x+x^2)\\
&~~~~~~+C(125-25x-5x^2+x^3)+D(25-10x+x^2)\\
1&=(C-A)x^3+(-5A+B-5C+D)x^2\\
&~~~~~~+(25A+10B-25C-10D)x+(125A+25B+125C+25D)
\end{align*}\]
This gives the system
\[\begin{cases}
C-A=0\\
-5A+B-5C+D=0\\
25A+10B-25C-10D=0\\
125A+25B+125C+25D=1
\end{cases}\]
I only slightly prefer this method over plugging in values that make some of the coefficients disappear. The downside is that's it much more tedious to solve the remaining equations...

Answer 19

I inherit @SithsAndGiggles stuff,
\[\frac{1}{(25-x^2)^2}=\frac{A}{5-x}+\frac{B}{(5-x)^2}+\frac{C}{5+x}+\frac{D}{(5+x)^2}\]
so that
\[1=A(5-x)(5+x)^2+B(5+x)^2+C(5+x)(5-x)^2+D(5-x)^2\]
if x =5, the right hand side is
the first term =0; the third term =0, the last term =0
so that\[ 1 = B (5+5)^2 \\B= \frac{1}{100}\]

Answer 20

if x =-5 , do the same, I have D =1/100

Answer 21

so, you just 2 A and C left
let x =4
1 = A(5-4)(5+4)^2 +1/100 (5+4)^2 +C(5+4)(5-4)^2+1/100(5-4)^2
1 = .....
let x =-4
do the same to have another equation, then solve them. It's not hard, right?

Answer 22

its tedious though. You're are so mart though too :D

Answer 23

Thank you, but I am not smart, just know the method.