# Particular Integral of arctan example

1. Aug 17, 2004

### relinquished™

I know for a fact that

$$\int \frac{(du)}{(a^2 + u^2)} = \frac{1}{a} \cdot arctan \frac{u}{a} + C$$

I was given the problem of solving the indefinite integral of

$$\int \frac{(dx)}{(2x^2 + 2x + 5)}$$

First, I multiplied the integral by (1/2) / (1/2) to eliminate the coefficient of the x^2 in the denominator, so now I am left with

$$\frac{1}{2} \cdot \int \frac{(dx)}{(x^2 + x + \frac{5}{2})}$$

Now, in completing the square of the denominator, I added $$\frac{1}{4} - \frac{1}{4}$$ (which is zero) so that the equation would look like this:

$$= \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{5}{2} + \frac{1}{4} - \frac{1}{4}}$$

Simplifying, I got:

$$= \frac{1}{2} \int \frac{dx}{x^2 + x + \frac{1}{4} + (\frac{5}{2} - \frac{1}{4})}$$

$$= \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + \frac{9}{4}}$$

$$= \frac{1}{2} \int \frac{dx}{(x + \frac{1}{2})^2 + ( \frac{3}{2})^2}$$

If we let $$(x + \frac{1}{2})^2 = u^2$$ and $$( \frac{3}{2})^2 = a^2$$ we now have the integrable form stated above, so

$$= \frac{1}{2} \int \frac{du}{u^2 + a^2}$$

$$= \frac{1}{2} \cdot \frac{2}{3} \cdot arctan \frac{u}{a} + C$$

$$= \frac{1}{3} \cdot arctan \frac{x^2 + \frac{1}{2}}{\frac{3}{2}} + C$$

$$= \frac{1}{3} \cdot arctan \frac{2(x^2 + \frac{1}{2})}{3} + C$$

My questions are:
(1) Is this the correct solution?
(2) If my solution is correct, how do you get the necessary constant in order to make the denominator of this problem a complete square? the (1/4 - 1/4) I added to the denominator just popped out of my mind. Is there any way to get it without resorting to trial and error?

thank you

2. Aug 17, 2004

### Muzza

1) No, there seems to be some confusion regarding the subsitution of variables, you've chosen (x + 1/2)^2 = u^2, this means that u = x + 1/2, not u = (x + 1/2)^2.

2) Suppose you wanted to write x^2 + bx + c in the form (x + p)^2 + q, i.e. you want to choose p, q such that x^2 + bx + c = (x + p)^2 + q = x^2 + 2px + p^2 + q. Comparing coefficients on both sides of the equation gives 2p = b and p^2 + q = c. So you can take p = b/2 and q = c - p^2 = c - (b/2)^2.

Last edited: Aug 17, 2004
3. Aug 17, 2004

### Zurtex

When I was doing this I found trying to complete all the quadratic equations I had by completing the square rather than any other method. Either actually solving them or just putting them in the form $a(x+b)^2 + c$, remember good algebra techniques greatly help calculus.

4. Aug 17, 2004

### relinquished™

um... since $$u^2 = ( x + \frac{1}{2} )^2$$ you can directly place this in the integrable form because the form needs a $$u^2$$,

but I understand the process of solving the quadratic eq'n or completing the square. thanx.

I was just wondering if the answer I got was correct.

5. Aug 17, 2004

### Zurtex

In the final two steps you have ended up replaces u with $x^2 + 1/2$, that was the only mistake I think.

6. Aug 18, 2004

### JonF

You also picked the wrong u originally. But then when you went to substitute you put in the right one. But later when you replaced u with x you went back to the wrong u.

7. Aug 18, 2004

### relinquished™

ahh.. ok, I see the mistake. Sorry. And the title should be Particular Integral that yields arctan. My mistake as well...

the u should be

$$u = x + \frac{1}{2}$$

and the real answer should be

$$= \frac{1}{3} \cdot arctan \frac{2(x + \frac{1}{2})}{3} + C$$

thanx again for the clarifications and replies. ^_^

Last edited: Aug 18, 2004
8. Aug 18, 2004

### JonF

I like your approach though; I probably would have used a ugly trig sub. I should probably memorize a few of the basic integral in tables.