Thread: Integration

1. You can write the given problem as
x^6-6x^5+10x^4-17x^3++8x^2-5x+1=A(x-1)^2(x^2+1)+B(x)(x1-1)^2(x^2+1)^2+C(x)(x-1)(x^2+1)^2+d(X)(x^2+1)+(Ex+f)(X)(x-1)^3(x^2+1)+(Gx+H)(x)(x-1)^3
Substituting x=0, you get A
Substituting x = 1 you get D
When you compare coefficients of same power of x, you get one equation.
In the given problem, there is no tern (x^2 +1) in the denominator, because E and F must be zero.
To find G, equate the coefficient of x^5 on both sides.
TO find H, equate the coefficient of x^4 on both sides.

2. Originally Posted by Debsta
OK so you'd set it up as:
A/x + B/(x-1) + C/(x-1)^2 + D/(x-1)^3 + (Ex+F)/(x^2+1) + (Gx+H)/(x^2+1)^2

Add it all up and equate your coefficients.
Remember to account for repeated factors on the bottom.
Also remember to put a polynomial of one less degree on the top.
Don't all the numerators have to be exactly one degree less than the denominators?

3. Originally Posted by Prove It
Don't all the numerators have to be exactly one degree less than the denominators?
I thought that is what I said. (Actually the degree on the numerator is one less than the degree of the "base" of the denominator.)

4. Originally Posted by sa-ri-ga-ma
In the given problem, there is no tern (x^2 +1) in the denominator, because E and F must be zero.
Is it possible that it's impossible to find without the already given answer?

without already knowing what the second rational expression is? I can only figure out A and D easily, and the rest seems really complicated to figure out

5. It is not impossible.
Expand the right hand side. Collect the terms of same power of x. Equate the coefficients of the same power of x from both sides. You will get seven equations. You have already found A and D. Find the rest of the constants.

6. Originally Posted by Debsta
I thought that is what I said. (Actually the degree on the numerator is one less than the degree of the "base" of the denominator.)
Then shouldn't the quadratic denominators have linear numerators, cubic denominators have quadratic denominators, and quartic denominators have cubic numerators? Because in your example none of the numerators go higher than a linear function...

7. Originally Posted by Prove It
Then shouldn't the quadratic denominators have linear numerators, cubic denominators have quadratic denominators, and quartic denominators have cubic numerators? Because in your example none of the numerators go higher than a linear function...
No the denominator (x-1)^3 for example will have a constant on the numerator. The degree on the numerator is one less than the BASE for a repeated factor. Similarly, (x^2+1) and (x^2+1)^2 will both have a linear numerator.

8. Ok Debsta i decided to just do it your way LOL, just to check whether my answer is right can you tell me the answer of the integral of that main challange?

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