# How would you Integrate this?

1. Nov 29, 2006

### lord12

integral of dx/x^(2)-6x +82

I know from completing the square you get the integral of dx/73 + (x-3)^(2). Then what?

2. Nov 29, 2006

### neutrino

Do the substitution x-3 = u and see if it turns into something a little more familiar.

3. Nov 29, 2006

### lord12

I just wanna know what you guys think as there is a discrepancy between my professors answer and my answer.

4. Nov 29, 2006

### neutrino

What's the discrepancy?

5. Nov 29, 2006

### lord12

my professor says that the final answer involves tan^(-1). I say that its (1/73)(u^(-2)du) = (1/73)(-(x-3)^(-1).

6. Nov 29, 2006

### neutrino

Your professor is right. Check your work again, the denominator is u^2 + 73

7. Nov 29, 2006

### AlphaNumeric

$$\int \frac{du}{u^{2}+A} = \int \frac{\sqrt{A}dy}{Ay^{2}+A} = \frac{1}{\sqrt{A}}\int \frac{dy}{y^{2}+1}$$

Combined with what others have said and what you should have in your notes you will be able to see the professor is right.

8. Nov 29, 2006

$$\int \frac{dx}{x^{2}-6x+82} \;$$

$$\int \frac{dx}{(x-3)^{2} + 73}$$

Use the fact that $$\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$$

So going back to the integral:

$$\int \frac{dx}{(x-3)^{2} + 73}$$, and $$u = x-3, \; du = dx$$

$$a^{2} = 73\Rightarrow a = \sqrt{73}$$

Thus $$\int \frac{dx}{(x-3)^{2} + 73} = \frac{1}{\sqrt{73}}\tan^{-1}\left(\frac{x-3}{\sqrt{73}}\right) + C$$

Last edited: Nov 29, 2006
9. Nov 29, 2006

### lord12

thanks people