1. ## integrating

how do you integrate

4u / (1+u)^2 (1+u^2) du

2. Originally Posted by jeph
how do you integrate

4u / (1+u)^2 (1+u^2) du
do you mean:

[4u/(1 + u)^2] * (1 + u^2) ??

3. oh no i meant

4u /[ (1+u)^2 * (1+u^2)] du

sorry didnt know it could have been interpreted like that o.o

4. Hello, jeph!

How do you integrate:
. . . . . . . . 4u
. . ∫ -------------------- du
. . . (1 + u)²(1 + u²)

You need Partial Fractions . . .

. . . . . 4u . . . . . . . . . . A . . . . . .B . . . . .Cu + D
-------------------- . = . ------- + ---------- + ---------
(1 + u)²(1 + u²) . . . . 1 + u . . (1 + u)² . . 1 + u²

5. Originally Posted by jeph
how do you integrate

4u / (1+u)^2 (1+u^2) du
Here

6. ## Re;

Re:

7. Originally Posted by qbkr21
Re:
Once you try to write the partial fractions, you have to respect this rule:
Denominator Containing Repeated Linear Factors

If a linear factor is repeated n times in the denominator, there will be n corresponding partial fractions with degree 1 to n.
You have to have that because you have to have the same polynomial as a denominator. Your expresion, does not have the same degree as the initial denominator.

8. ## Re:

Re:

9. Originally Posted by qbkr21
Re:
yes, that's correct

10. ## Re:

Re:

11. Originally Posted by qbkr21
Re:
we can find C immediately by plugging in x = -1, but for the rest, it seems you would have to multiply out to get them

12. ## Re:

RE:

Yea I found c to be equal to (-1/4)

I will post my work as I progress along on this problem tonight!

Thanks!

13. Originally Posted by Soroban
Hello, jeph!

You need Partial Fractions . . .

. . . . . 4u . . . . . . . . . . A . . . . . .B . . . . .Cu + D
-------------------- . = . ------- + ---------- + ---------
(1 + u)²(1 + u²) . . . . 1 + u . . (1 + u)² . . 1 + u²
Before to apply partial fractions, I'd check first that $(1+u)^2-(1+u^2)=2u$, which frees to us of the previous mechanism.

Regards