# Techniques of integration, Partial Fractions problem.

1. Aug 26, 2008

### afcwestwarrior

1. The problem statement, all variables and given/known data

∫(2t)/(t-3)^2

the integral is 2 to 0

ok does it = A/ t-3 + B/(t-3)^2

I'm not sure if you break up (t-3)^2

2. Aug 26, 2008

### konthelion

Yep, $$\frac{2t}{ (t-3)^2} = \frac{A}{t-3} + \frac{B}{(t-3)^2}$$.

From here, just multiply both sides by $$(t-3)^2$$

Edit: Sorry, I used the variable 'x' instead of 't'.

3. Aug 26, 2008

### afcwestwarrior

It's ok.

4. Aug 26, 2008

### afcwestwarrior

So would it be 2t= A(t-3)+ B

=At-3A+B
A(t) - (3A+B)

A=2
3A+B=0

Is this correct.

5. Aug 26, 2008

### konthelion

You should get
A = 2 and B - 3A = 0

6. Aug 26, 2008

### afcwestwarrior

you distributed the negative to make it B-3A

7. Aug 26, 2008

### afcwestwarrior

A=2
and B=6

then would it be 2/ t-2 + 6/ (t-2)^2

skipping u substitution part

= 2 ln[t-2] + 6 ln [(t-2)^2

is that correct

8. Aug 26, 2008

### konthelion

You have a typo. It's a 3 not a 2.

$$\int \frac{2t}{ (t-3)^2}dt =\int \frac{2}{t-3}dt + \int \frac{6}{(t-3)^2}dt$$

Also, you did not integrate the right term correctly, it's a degree of "-2" not "-1"

9. Aug 26, 2008

### afcwestwarrior

A=3 is what your saying, I'm not following. Where's my mistake

10. Aug 26, 2008

### konthelion

It is (t-3) not (t-2)

11. Aug 27, 2008

### afcwestwarrior

oh ok I understand. so would it be 2 Ln [t-3] + 6 ln[(t-3)^2]

12. Aug 27, 2008

### Defennder

This part is incorrect. What is the anti-derivative of $$\frac{1}{(t-3)^2}$$?

13. Aug 27, 2008

### HallsofIvy

The simple way to solve 2t= A(t-3)+ B, for all t, is to let t= 3 first: 2(3)= A(0)+ B so B= 6. Then, if, say, you let t= 0, 0= -3A+ 6 so 3A= 6 and A= 2.