When does $\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$ converge? I am required to find the values of $C$ for which the integral
$$\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$$ converges.
I know by experimentation that it converges when $C=3$. I am, however, unable to show this in a rigorous way.
I get stuck when I need to evaluate $$\lim_{x\rightarrow\infty}(\frac{7}{2}\ln(x^2+1)-\frac{7C}{3}ln(3x+1))$$
Any help would be appreciated.

Taking the hint from DonAntonio, I wrote the following solution. Let me know if there is anything wrong with the way I expanded DonAntonio's solution. 
$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$
When $C=3$
$$7\cdot\,\int_0^{\infty}\frac{(3-3)x^2+x-3}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{x-3}{(x^2+1)(3x+1)}dx<\int_0^{\infty}\frac{x}{x^3}dx = \int_0^{\infty}\frac{1}{x^2} dx$$
Which is convergent.
When $C\neq3$
$$7\cdot\,\int_0^{\infty}\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}+7\cdot\,\int_0^{\infty}\frac{x-C}{(x^2+1)(3x+1)}dx$$
$$=|7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}dx+K| > \int_0^{\infty}\frac{x^2}{x^4} dx = \int_0^{\infty}\frac{1}{x^2} $$
For some constant $K$
Which is divergent.

Hints:
$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$
Now, if $\;C\neq3\;$ then comparison with some multiple of $\;\frac1x\;$ will give you divergence...
