Question

Integrate x^2/((x-1) (x+1) (x+1)) dx
Can someone please help me?

Answer 1

partial fractions

Answer 2

How do I do the plug in method? Where I plug in a value to solve for the variables.

Answer 3

I got to:
A(x+1)^2 +B(x+1)(x-1) + (Cx+D)(x-1)=x^2

Answer 4

https://www.youtube.com/watch?v=_YMG5mQTAOg This guy is really good at explaining it.

Answer 5

It's how I learned

Answer 6

Take the integral:
integral x^2/((x-1) (x+1)^2) dx
For the integrand x^2/((x-1) (x+1)^2), use partial fractions:
= integral (3/(4 (x+1))-1/(2 (x+1)^2)+1/(4 (x-1))) dx
Integrate the sum term by term and factor out constants:
= 3/4 integral 1/(x+1) dx-1/2 integral 1/(x+1)^2 dx+1/4 integral 1/(x-1) dx
For the integrand 1/(x+1), substitute u = x+1 and du = dx:
= 3/4 integral 1/u du-1/2 integral 1/(x+1)^2 dx+1/4 integral 1/(x-1) dx
The integral of 1/u is log(u):
= (3 log(u))/4-1/2 integral 1/(x+1)^2 dx+1/4 integral 1/(x-1) dx
For the integrand 1/(x+1)^2, substitute s = x+1 and ds = dx:
= (3 log(u))/4-1/2 integral 1/s^2 ds+1/4 integral 1/(x-1) dx
The integral of 1/s^2 is -1/s:
= 1/(2 s)+(3 log(u))/4+1/4 integral 1/(x-1) dx
For the integrand 1/(x-1), substitute p = x-1 and dp = dx:
= 1/(2 s)+(3 log(u))/4+1/4 integral 1/p dp
The integral of 1/p is log(p):
= (log(p))/4+1/(2 s)+(3 log(u))/4+constant
Substitute back for p = x-1:
= (3 s log(u)+s log(x-1)+2)/(4 s)+constant
Substitute back for s = x+1:
= (3 (x+1) log(u)+(x+1) log(x-1)+2)/(4 (x+1))+constant
Substitute back for u = x+1:
= ((x+1) log(x-1)+3 (x+1) log(x+1)+2)/(4 (x+1))+constant
Which is equal to:
Answer: = 1/4 (2/(x+1)+log(x-1)+3 log(x+1))+constant

Answer 7

sorry let me clean that up

Answer 8

Wait....
How did you get the 3/4, 1/2 and 1/4?

Answer 9

That's my question....

Answer 10

I think I understand now...

Answer 11

Thanks guys!!

Answer 12

Sure thing! good luck!

Answer 13

sorry i wasn't too helpful lol