Calculus
posted by Cesar .
Evaluate the integral
INT(x^4+6x^3+5x^2+8)/(x^2+6x+5)dx
Please help.

The integrand is:
f(x) = (x^4+6x^3+5x^2+8)/(x^2+6x+5)
The denominator can be factored as:
x^2+6x+5 = (x+1)(x+5)
This means that f(x) has singularities at x = 1 and x = 5, if the numerator isn't zero there. To find the partial fraction expansion, you can simply calculate the singular behavior near these points, subtract from f(x), leving you with a rational function without any singularities, which is therefore a polynomial (which you can easily find).
This has the advantage of not having to solve equations to find the partial fraction expansion. To find the bahavior near x = 1, you expand the function in powers of (x+1), only the leading term is singular
f(x) = 2/(x+1) + nonsingular terms
The expnsion around x = 5 yields:
f(x) = 2/(x+5) + nonsingular terms
If we subtract the singular terms from
f(x) we will be left with a rational functuon without any singularities, which is therefore a polynomial.
Let's see what we get:
f(x)  2/(x+1) + 2/(x+5) =
[x^4+6x^3+5x^2+8  2(x+5) + 2 (x+1)]/
[(x+1)(x+5)] =
[x^4 + 6 x^3 + 5 x^2]/[(x+1)(x+5)] =
x^2 (x^2 + 6 x + 5)/[(x+1)(x+5)] =
x^2
So, we have:
f(x)  2/(x+1) + 2/(x+5) = x^2 >
f(x) = x^2 + 2/(x+1)  2/(x+5)
The integral is thus:
1/3 x^3 + 2 Logx+1  2 Logx+5 + c 
(x^4+6x^3+5x^2+8)/(x^2+6x+5)
= x^2 + 8/(x^2 + 6x+5)
= x^2 + 8/((x+1)(x+5))
let 8/((x+1)(x+5) = A/(x+5) + B/(x+1)
8 = A(x+1) + B(x+5)
let x = 1
8 = 4B > B = 2
let x = 5
8 = 4A
A = 2
so ∫(x^4+6x^3+5x^2+8)/(x^2+6x+5) dx
= ∫ (x^2  2/(x+5) + 2(x+1) dx
= (1/3)x^3  2ln(x+5) + 2ln(x+1) + c 
Did not realize that Count Iblis had already answered this.
I guess it helps to refresh the page before posting.
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