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September 1, 2014

September 1, 2014

Posted by **Cesar** on Tuesday, September 20, 2011 at 12:28pm.

INT(x^4+6x^3+5x^2+8)/(x^2+6x+5)dx

Please help.

- Calculus -
**Count Iblis**, Tuesday, September 20, 2011 at 1:00pmThe 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) + non-singular terms

The expnsion around x = -5 yields:

f(x) = -2/(x+5) + non-singular 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 Log|x+1| - 2 Log|x+5| + c

- Calculus -
**Reiny**, Tuesday, September 20, 2011 at 1:18pm(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

- Calculus -
**Reiny**, Tuesday, September 20, 2011 at 1:20pmDid not realize that Count Iblis had already answered this.

I guess it helps to refresh the page before posting.

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