# Thread: Integral ( complete the square )

1. ## Integral ( complete the square )

I have a problem in my book where i have an integral and i have to use complete the square if necessary. I understand how to do the complete the square. However, this problem don't need it but its stumping me. I am not sure how the solutions manual got the second step. It split it up into two integrals and i cant see how it came to that conclusion?

$\int \frac {2x}{x^2+6x+13} dx$

The next step is

$\int \frac {2x + 6} {x^2 + 6x + 13} dx - 6 \int \frac {1}{x^2+6x+13} dx$

How are they splitting up the integral and where is the 2x +6 in the first integral and -6 in the second coming from?

The next step( which i left out) looks like they use complete the square on the second integral.

Thank you so much!!!!!

2. ## Re: Integral ( complete the square )

$\int~\dfrac{2x}{x^2+6x+13} dx$

$=\int~\dfrac{2x+(6-6)}{x^2+6x+13} dx$

$=\int~\dfrac{2x+6}{x^2+6x+13}dx-\int~\dfrac{6}{x^2+6x+13} dx$

$=\int~\dfrac{2x+6}{x^2+6x+13}dx-6\int~\dfrac{1}{x^2+6x+13} dx$

Then, I think you probably would complete the square on the second integral.

3. ## Re: Integral ( complete the square )

Why did they choose the 6 though? Why couldn't i have used any number?

4. ## Re: Integral ( complete the square )

Because using the 6 makes the first integral easy to solve. This is because it is now of the form:

$\int~\dfrac{f'(x)}{f(x)} dx = ln(f(x))+ C$

5. ## Re: Integral ( complete the square )

I understand your formula. However, i am still confused on how i would have known to use a 6? You said it make it easier but how would i have known that? Why couldnt i have used a 3 instead? Im sorry but im just baffled.

6. ## Re: Integral ( complete the square )

I understand your confusion. When you see a fraction that you have to integrate, there are a few things to try. It really is a case of "manipulate, guess and check."

-Partial fractions. Here this isn't possible because the denominator doesn't factor.
-Substitutions. Here, there *might* be a substitution that will make life easier but I certainly wouldn't spot it easily.
If neither of these will work, then check whether you can somehow make something of the form $\frac{f'(x)}{f(x)}$ - what my teacher used to call a "related integral" for some reason.

Start with the denominator. Here, this is $x^2+6x+13$. If we differentiate this, we get $2x+6$. If this is our numerator, then great! We have something of the form $\frac{f'(x)}{f(x)}$. But it isn't.

$\frac{2x+6}{x^2+6x+13}$ is of the form $\frac{f'(x)}{f(x)}$ but $\frac{2x}{x^2+6x+13}$ is not.

We cannot easily integrate

$\frac{2x}{x^2+6x+13}$

...so we make it into:

$\frac{2x+6}{x^2+6x+13}-\frac{6}{x^2+6x+13}$ which we can then integrate. We don't try '3' because this does not relate to the denominator. We get the '6' from differentiating the denominator.

7. ## Re: Integral ( complete the square )

Perfect! That explained it perfectly. The U prime over U should have been a clue to me. Thank you so much!!!!