Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm to integrate $\int\frac{x}{x^2+6x+13}dx$

But I'm finding it impossible to do anything with it.

Since $x^2+6x+13$ is an irreducible quadratic factor, and the only factor it means that the partial fraction should be in the form...

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

Hence $x = (Ax+B)(x^2+6x+13)$

$=>x= Ax^3+(6A+B)x^2+(13A+6B)x+13B$

Equating coefficients mean that A =0, 6A+B = 0, 13A+6B = 1 and 13B = 0.

Obviously this doesn't hold. Since 13A+6B apparently = 1 despite A and B both coming out as zero.

The other way I did it is

Let x = 0, hence 0 = 13B, therefore B = 0. Let x = 1, hence 1 = A + 6A + 13A = 20A, therefore A = $\frac{1}{20}$

But that means $\int\frac{x}{x^2+6x+13}dx$ = $\int\frac{x}{20(x^2+6x+13)}dx$, which isn't possible unless x is always 0. And it doesn't help me integrate it.

Quite clearly I've done something wrong something along the lines, but I have no idea what?

Any pointers?

share|improve this question
    
Of course $\int\frac{x}{x^2+6x+13}dx$ is already in "partial fraction" form. Next (as DonAntonio says) complete the square. –  GEdgar Mar 9 at 14:17

3 Answers 3

Hints:

$$\frac x{x^2+6x+13}=\frac12\frac{2x+6}{x^2+6x+13}-\frac3{(x+3)^2+4}=$$

$$\frac12\frac{(x^2+6x+13)'}{x^2+6x+13}-\frac32\frac{\left(\frac{x+3}2\right)'}{1+\left(\frac{x+3}2\right)^2}$$

share|improve this answer
2  
Just what I was going to post!! +1 –  amWhy Mar 9 at 14:10

Actually, it is a simple partial fraction and your mistake is $$x = (Ax+B)\color{red}{(x^2+6x+13)}$$ the red part should be deleted and so $$x = (Ax+B)\implies A=1,B=0$$ For the integration, you have $$\frac{1}{2}\int\frac{(2x+6)-6}{x^2+6x+13}dx\\\frac{1}{2}\int\frac{(2x+6)}{x^2+6x+13}-\frac{6}{(x+3)^2+4}dx$$

share|improve this answer
    
Silly me! Though that just brings us back to the original function which is still difficult to integrate. :\ –  Wolff Mar 9 at 14:11
    
@Wolff is it easy now –  Semsem Mar 9 at 14:13
    
I was able to integrate the first part to get $\frac{1}{2}ln|x^2+6x+13|$ but I'm still stuck on integrating $1/2 . \int 6/((x+3)^2+4)dx$ :S –  Wolff Mar 9 at 14:28
    
@Wolff Think inverse trigonometric functions. –  Eric Thoma Mar 9 at 17:47
    
@Wolff it should be $\frac{6}{\sqrt4}\tan^{-1}(\frac{x+3}{\sqrt4})$ –  Semsem Mar 9 at 20:05

First, force the apparition of the derivative of the denominator of the fraction in the numerator.

Then you have to deal with $$ \int \frac{dx}{x^2 + 6x + 13} $$and do this write $$ x^2 + 6x + 13 = (x+3)^2 + 2^2 $$

share|improve this answer
1  
I'm confused as to what you did with the numerator, x? –  Wolff Mar 9 at 14:10
    
write $x = \frac{2x + 6}2 - 3$. –  mookid Mar 9 at 14:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.