partial integral?

Question

x^2 / (x+1)(x^2+2x+2) equals x^2/ (x+1)[(x+1)^2] + 1

how do i break this into A/x + B/x + C/x with that + 1 being there? Thanks!

x^2 + 2x + 2 = (x + 1)^2 + 1. I d

Answer 1

x² / [(x+1) (x²+2x+2)] cannot be expressed in the form A/x + B/x + C/x, but like this:

x² / [(x+1) (x²+2x+2)] = A /(x+1) + (Bx+C) / (x²+2x+2)

So,

x² = A (x²+2x+2) + (BX +C)(x+1)
regrouping:
x² = (A+B)x² + (2A+B+C)x + (2A+C)
Then
1 = A+B
0=2A+B+C
0=2A+C
Solving A,B and C:
A=1, B=0, C=-2

x² / [(x+1) (x²+2x+2)] = 1 / (x+1) - 2 / (x²+2x+2)

But if you want to integrate:
∫x² / [(x+1) (x²+2x+2)] dx
Before to integrate, we can see that we need to re-write (x²+2x+2)
as:
(1+ (x+1)²)

= ∫[1 / (x+1) - 2 / (1+ (x+1)²)] dx

= ln|x+1| - 2arctan(x+1) +c

Answer 2

Benjamin, YOUR premise is wrong. He is right. The two are NOT equal if you look carefully. He is basically saying that
x^2+2x+2=(x+1)^2+1
He is simply completing the square.

Now Les, what you have done is not wrong but it is a step in the wrong direction, you don't want to complete the square like that.

Leave it like, x^2/(x+1)(x^2+2x+2) and decompose it into
A/(x+1) + (Bx+C)/(x^2+2x+2)

So now we have that
x^2 = Ax^2+2Ax+2A+Bx^2+Bx+Cx+C

which gives us that
A+B=1
2A+B+C=0
2A+C=0

If we solve the system, we get that
A=1
B=0
C=-2

Which means that the integral is now
1/(x+1) - 2/(x^2+2x+2)

take the integral of the first one and we get ln(x+1)

Now for the second one you complete the square and use tangent substitution and we get

ln(x+1) - 2arctan(x+1)+C

Answer 3

Your premise is incorrect--the two are not equal. How can something equal itself +1 (assuming you're not doing modular arithmetic)?

Also, I think you are looking for something of the form

A/(X+1)+B/(x+1)^2+C/(X+1)^3

That is how partial fraction decompositions work.