1. another integration problem

(x-2)/(x^2+x^4) dy
I tried this way, but the result didn't seem right.
(AX+B)(X^2+1)+(C)/(X)+(D)/(X^2)

2. Originally Posted by questionboy
(x-2)/(x^2+x^4) dy
I tried this way, but the result didn't seem right.
(AX+B)(X^2+1)+(C)/(X)+(D)/(X^2)
Your partial fraction decomposition is correct. Continue using it.

3. really?
then i have
(-x+2)/(x^2+1)+1/x+(-2/x^2) dx

then
{ln|x|-2ln|x^2|} +{(-x+2)/(x^2+1) dx}
Im stuck here

i tried this
{(-x+2)/(x^2+1) dx}
a=1
x=tany
dx=secy^2 dy
[(-tany+2)*secy^2]/ (secy^2)
=-tany+2
i don't know what to do now

4. Originally Posted by questionboy
really?
then i have
(-x+2)/(x^2+1)+1/x+(-2/x^2) dx

then
{ln|x|-2ln|x^2|} +{(-x+2)/(x^2+1) dx}
Im stuck here

i tried this
{(-x+2)/(x^2+1) dx}
a=1
x=tany
dx=secy^2 dy
[(-tany+2)*secy^2]/ (secy^2)
=-tany+2
i don't know what to do now
All integrals are standard forms. Omitting the C's:

$\displaystyle \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = - \frac{1}{x}$.

$\displaystyle \int \frac{-x + 2}{x^2 + 1} = - \int \frac{x}{x^2 + 1} \, dx + \int \frac{2}{x^2 + 1} \, dx = - \frac{1}{2} \ln |x^2 + 2| + 2 \tan^{-1} x$.

5. Thank you very much.Master