# Thread: Inergration With Partial Fractions

1. ## Inergration With Partial Fractions

$\int \frac {x-4}{x^2+2x+5}dx$

Answer: $\frac{1}{2}ln(x^2+2x+5)+\frac{3}{2}\arctan(\frac{x +1}{2})+C$

Well, this one I'm not sure how to even start. I can't factor out the bottom, and making it a perfect square and separating the fractions makes to a lot more complicated than i think it needs to be. Can anyone give me any tips on how to Start/Solve this problem?

Thanks

2. Notice how the top is the derivative of the bottom. This gives you your first time: $\int \frac {f'(z)}{z} dz$ is one of the result you should have at your fingertips.

No it's not, I'm talking rubbish. Sorry, forget I spoke ...

Okay, so make the top (x + 1 - 5) and get 2 integrals, one of x+1 / x^2 + 2x + 5, and one of -5 / x^2 + 2x + 5

and the first part drops out.

3. $\int \frac {x-4}{x^2+2x+5}dx$

Multiply and divide by 2: $\frac{1}{2}\int \frac {2x-8}{x^2+2x+5}dx$

Separate it so that the derivative of the denominator is at the top: $\frac{1}{2}\int \frac {2x+2}{x^2+2x+5}dx - \frac{1}{2}\int \frac{10}{x^2+2x+5}dx$

Can you do it now?

P.S. Matt, you're not talking rubbish at all. :P

EDIT: I see you edited your post.

4. Originally Posted by mortalapeman
$\int \frac {x-4}{x^2+2x+5}dx$

Answer: $\frac{1}{2}ln(x^2+2x+5)+\frac{3}{2}\arctan(\frac{x +1}{2})+C$

Well, this one I'm not sure how to even start. I can't factor out the bottom, and making it a perfect square and separating the fractions makes to a lot more complicated than i think it needs to be. Can anyone give me any tips on how to Start/Solve this problem?

Thanks
BTW, the second part of your answer is not correct. :O

5. Originally Posted by Chop Suey
$\int \frac {x-4}{x^2+2x+5}dx$

Multiply and divide by 2: $\frac{1}{2}\int \frac {2x-8}{x^2+2x+5}dx$

Separate it so that the derivative of the denominator is at the top: $\frac{1}{2}\int \frac {2x+2}{x^2+2x+5}dx - \frac{1}{2}\int \frac{10}{x^2+2x+5}dx$

Can you do it now?

P.S. Matt, you're not talking rubbish at all. :P

EDIT: I see you edited your post.
Our teacher hasn't taught us what having the derivative of the denominator in the numerator means. Like i can see what to do with the constant in the numerator, but don't know how to work the other one 0o

6. Originally Posted by Chop Suey
BTW, the second part of your answer is not correct. :O
Hehe talk to the author of my math book, that happens to be an odd numbered prob. That's the only reason i even have an answer.

7. Originally Posted by mortalapeman
Our teacher hasn't taught us what having the derivative of the denominator in the numerator means. Like i can see what to do with the constant in the numerator, but don't know how to work the other one 0o
You can't solve the second one? Just complete the square and it becomes a simple arctan integral.

And what part of this you don't understand? It's just algebraic manipulation to make things fit.

if you have an integral in this form:
$\int \frac{f'(u)}{f(u)} du$

Then you can simply substitute:
$z = f(u)$

$dz = f'(u) du$

And then replace:
$\int \frac{1}{z} dz = \ln|z| + C$

$\therefore \int \frac{f'(u)}{f(u)} du = \ln|f(u)| + C$

8. Ah i knew that, i just didn't recognize it. I've never dealt with derivatives that were more than 1 term, so i didn't think of it that way. Thank you, i'll have to get onto my professor about this one