Evaluate the integral from $0$ to $\infty$

Question

Consider the improper integral

$$\int_0^\infty \frac{x}{x^4 +9}dx$$

So first I take the limit from b to $\infty$ to fix the integral

$$\lim_{b\to \infty} \int_0^b \frac{x}{x^4 +9}dx$$

now here is where I'm stuck, do I use trig sub to solve it or use a U-sub to get an arctan?

Answer 1

Substitute $$u=x^2$$ then we have $$du=2xdx$$

Answer 2

Observe that

$$x^4+9=(x^2+\sqrt6\,x+3)(x^2-\sqrt6\,x+3)$$

so now do partial fractions...

Answer 3

Hint:

$$\frac{x\,dx}{x^4+9}=\frac32\frac19\frac{d(\frac{x^2}3)}{(\frac{x^2}3)^2+1}.$$