Find the integral of x/(x^2 +4x +13)

[mat331760298]

1. Find the integral of x/(x^2 +4x +13)

I'm not sure if I need to complete the square or what. I really don't know where to begin.

 

[eumyang]

Try rewriting the integrand as a difference of two fractions.

 

[HallsofIvy]

Yes, start by completing the square. That will reduce it to something like $(x- a)^2+ b$. Then let u= x- a so that denominator becomes $u^2+ b$ and the numerator becomes x= u+ a:
$$\int \frac{x}{x^2+ 4x+ 13}= \int \frac{u}{u^2+ b}du+ \int\frac{a}{u^2+ b}du$$
The first integral can be done by the substitution $v= u^2+ b$ and the second is an arctangent.

 

[hunt_mat]

First however note that:
$$\frac{x}{x^{2}+4x+13}=\frac{1}{2}\frac{2x+4}{x^{2}+4x+13}-\frac{2}{x^{2}+4x+13}$$
When integrating this expression, the first term is of the form f'(x)/f(x) which integrates to log(f(x)). Complte the square for the second term.