Is this fraction even possible to put into partial fractions?

Question

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?

Answer 1

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 $$\frac{1}{2}\int\frac{(2x+6)-6}{x^2+6x+13}dx\\\frac{1}{2}\int\frac{(2x+6)}{x^2+6x+13}-\frac{6}{(x+3)^2+4}dx$$

Answer 2

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 $$ x^2 + 6x + 13 = (x+3)^2 + 2^2 $$

Answer 3

Hints:

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

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