Integration Problem which requires some serious help.

Hi, I have been studying diffentation over the last few months and i have recently moved on to Integration and i need some help. (I am struggling).

Can any body out there take me through the following example question to help me grasp the fundamentals on this tricky subject. Here goes

A= 2f + 1/(f+3) (f-3)
(i) Express A as a sum of partial fractions.
(ii) Intergrate A with respect to f.

Any help would be greatly appreciated.

Many Thanks
Paul

(i) You have to break down the term 1/(f + 3)(f - 3) (I hope the whole (2f + 1) is not in the numerator; the solution will look quite different if it is).

1/(f + 3)(f - 3) = A/(f + 3) + B/(f - 3)...............multiply by (f + 3)(f - 3)

1 = A(f - 3) + B(f + 3)..........A + B = 0; -3A + 3B = 1........B = 1/6, A = - 1/6

so now A = 2f + (1/6)/(f - 3) - (1/6)/(f + 3)

Now you can integrate term by term to get f^2 + 1/6(ln(f - 3) - ln(f + 3)) + C (constant of integration). If the 2f + 1 is in the numerator, the partial fractions look like

(2f + 1)/(f + 3)(f - 3) = A/(f + 3) + B/(f - 3)

2f + 1 = A(f - 3) + B(f + 3).......A + B = 2.......-3A + 3B = 1........B = 7/6 and A = 5/6

Then when you integrate you get (1/6)(5ln(f + 3) + 7ln(f - 3)) + C

A = (2f + 1) / [(f + 3)(f - 3)]

(2f + 1) / [(f + 3)(f - 3)] = a/(f + 3) + b/(f - 3)
2f + 1 = a(f - 3) + b(f + 3)
2f + 1 = af - 3a + bf + 3b
2f + 1 = (a + b)f + 3(b - a)

a + b = 2
3(b - a) = 1
----------------
b - a = 1/3
b = 1/3 + a

a + b = 2
a + (1/3 + a) = 2
2a + 1/3 = 2
2a = 5/3
a = 5/6

b = 1/3 + a
b = 1/3 + 5/6
b = 7/6

A = (2f + 1) / [(f + 3)(f - 3)]
A = a/(f + 3) + b/(f - 3)
A = (5/6)/(f + 3) + (7/6)/(f - 3)]
A = 1/6 * [5/(f + 3) + 7/(f - 3)]

You should be able to integrate both terms and get 2 natural logs.

ii) f^2 + ln(f^2-9) + c