# Aggravating Calculus problem; proceed at your own risk :P

1. Aug 29, 2009

1. $$\int \frac{x}{(x-3)^{2}+1}$$
which was had by completing the square integrate
$$\int \frac{x}{x^{2}-6x+10}$$

$$u = x$$
$$du = 1$$
$$dv=\frac{1}{(x-3)^{2}+1) }$$
$$v=arctan(x-3)$$

3. The attempt at a solution
This problem was presented in a refresher sheet for an engineering course I'm taking this semester. I completed many long and tedious integrations but seem to run into a problem on only this one from the entire sheet. I could easily just forget this problem and use the arctan solution in the integral tables which I inadvertently memorized while trying to solve this problem. If I attempt to integrate by parts from the beginning using the completed square I get the following; which is where I run into a problem. If someone could work out a proper solution I would be eternally grateful even if you just dump it in without using latex.

$$x*arctan(x-3) - \int arctan(x-3)$$

Integrating arctan when only x or x with a coefficient is easily done by integrating by parts. However I seem to repeat the initial problem when I try to integrate it regardless. The whole difficulty seems to lie with having u=(x-3) rather than (x) or (nx)....

Last edited: Aug 29, 2009
2. Aug 29, 2009

### arildno

You may proceed as follows:
$$\int\frac{x}{(x-3)^{2}+1}dx=\int\frac{(x-3)+3}{(x-3)^{2}+1}dx=\int\frac{(x-3)}{(x-3)^{2}+1}dx+\int\frac{3}{(x-3)^{2}+1}dx$$

Both of these two integrals are easy to compute.

3. Aug 29, 2009

you sir are made of win sauce!

I will post my final answer in a moment for anyone else who stumbles upon this problem!

4. Aug 29, 2009

Wow thanks for that trick, I will make sure to utilize it fully in the future. Silly me!

$$=\frac{1}{2}ln((x-3)^{2}+1) + 3arctan(x-3)$$

5. Aug 29, 2009

And I'm happy to say that Mathematica agrees with me :)
Thanks, you really rock!

6. Aug 29, 2009

### arildno

Okay, I guess I can't complain for lack of compliments...:shy:

7. Sep 14, 2009

$$C_{T}=\int^{0}_{1}\frac{\sigma}{2}r^{2}c_{l}dr$$

$$\sigma=\frac{Nc}{\pi R}$$
this assumes constant chord (i need to revise this for blade taper)
$$c_{l}=a\alpha$$
$$C_{p}=\int \lambda dC_{T}+\int^{0}_{1}\frac{\sigma}{2}r^{3}c_{d}dr$$
$$\lambda=\frac{\sigma a}{16}[\sqrt{1+\frac{32}{\sigma a}\theta r}-1]$$