# Using the u of substitution

1. Apr 12, 2013

### nick.martinez

∫1/(10p-p^2)dp

i tried using the u of substitution but for some reason I am unable to isolate dp and get an equation in terms of du which i could then plug into the integral and take the antiderivative.

2. Apr 12, 2013

### CompuChip

Since there is a p2 in the denominator, the first thing that comes to mind is
$$\int \frac{1}{1 + p^2} \, dp = \arctan(p)$$

So I would try completing the square, i.e. write it as
$$k \cdot \int \frac{1}{1 + u^2} \, du$$
where u = (p + a) / b.

3. Apr 12, 2013

### nick.martinez

not quite sure where you got u=(p+a)/b

4. Apr 12, 2013

### nick.martinez

do you mean i should factor out a 10 from denominator which would give me:

(1/10)∫1/(p-(p^2)/10)dp

5. Apr 12, 2013

### CompuChip

No, start by writing 10p - p2 as (p - a)2 + b.

6. Apr 12, 2013

### CompuChip

By the way, I was assuming that you know the arctan integral. If not, you could try splitting fractions, i.e. writing
$$\frac{10p - p^2} = \frac{A}{10p} - \frac{B}{p^2}$$

7. Apr 12, 2013

### nick.martinez

I would like get this into a Ln function instead of using arctan

8. Apr 12, 2013

### nick.martinez

are you using a partial fraction decomp in this case. if so that would be my preferred choice at getting to a solution.

9. Apr 12, 2013

### nick.martinez

when i find the perfect square how do i find what the perferct square would be. I dont even know where to start. i punched the problem into wolfram and it completed the square of denom by turning

1/(10p-p^2) which i guess is equal to 1/(25-(x-5)^2), how do i do this?

10. Apr 12, 2013

### CompuChip

OK in that case first try with the partial fraction decomposition, once you have arrived at the solution I will show you the method I was thinking of initially for completeness.

11. Apr 12, 2013

### Staff: Mentor

I'm sure you mean
$$\frac{1}{p(10 - p)} = \frac{A}{p} + \frac{B}{10 - p}$$