Ok, so do it. Type up the next step.
Any integer can go into something! If you mean that no number goes into any of them, I would agree with you. That's why I picked primed numbers. However, you don't need to do that; instead, you need to find the least common multiple of all the denominators. How do you do that?
not sure, its been about 15 year since I did basic maths. I think the best thing would be to start backwards. I know how to work out the partial fractions for 2 and 3 parts. But not sure how they were derived. for example I know from memory that A/(S-1) + B/(S-1)^2 + C(S+3) becomes A(S-1)(S+3) +B(S+3) + C(S-1)^2. I dont know why, I just memorised it.
Right. So you added this way, correct?
$\displaystyle \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{3}{6}+\dfrac{2}{6 }=\dfrac{3+2}{6}=\dfrac{5}{6}?$
Now, add these fractions together:
$\displaystyle \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}.$
What do you get? Show all your steps just like I did above with the 1/2 + 1/3.
In general, you'd be right. However, in this case, the left-hand fraction doesn't factor over the reals (the discriminant $\displaystyle b^{2}-4ac<0$.) Therefore, you have to view the entire denominator as a factor. In other words, the entire denominator of the left-hand fraction acts like one of the prime numbers in the denominators of the fractions in post # 8. How does that change things for you?