# Cal II Integration

1. May 7, 2007

### CallingAllCars7

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

find the following integral:

(t+7)/(5-t)^(1/2)

2. Relevant equations
maybe U*v'= uv-[ u'-v] or substitution

3. The attempt at a solution
Ive been trying substitution
u= (5-t)^(1/2)

t= 5-u^2
dt= -2u

((5-u^2)+7)/u )*-2u(du)

I get stuck here and I'm not even sure if I'm on the right track. I've seen the solution but I just cant get the answer.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 7, 2007

### Mindscrape

Have you tried integration by parts?

3. May 7, 2007

### Curious3141

Rearrange it algebraically first and it becomes a doddle.

$$\frac{t+7}{\sqrt{5-t}} = -\frac{(5-t)-12}{\sqrt{5-t}} = -{{(5-t)}^{\frac{1}{2}} + 12{({5-t})^{-\frac{1}{2}}$$

Last edited: May 7, 2007
4. May 8, 2007

### HallsofIvy

Why would you get stuck here? 5- u2+ 7= 12- u^2 and the u in the denominator cancels the u in "-2udu". Simple algebra gives
(-23+ 2u^2)du

5. May 9, 2007

### CallingAllCars7

its just that the solution looks nothing like that :

its -2t(5-t)^(1/2)- (4/3)(5-t)^(3/2)-14(5-t)^(1/2)+c

I did also try integration by parts with another different answer.:grumpy:

6. May 9, 2007

### Gib Z

Personally I would not have been clever enough to arrange it as Curious3141 did, but Thats the way I would like to have done it :)

I would have split it into 2 integrals, by splitting up the numerator. Then I would have tried t= 5 sin x, dt= 5 cos x dx on the first one, and u=5-t on the second. It gets you there, but Curious did it much better.