How to integrate problem I've tried u substitution and partial fraction doesn't seem to work either.
$\int \frac{1}{x^2+3}$
Can I take $\int \frac{1}{x^2+\sqrt(3)^2}$ in order to use $x=a\tan\theta$

We are asked to evaluate $ \int \frac{1}{x^2+3}dx$. We need to know $\int \frac{1}{1+x^2}dx= \arctan(x)+C$ to answer this question. 
We need to modify the denominator $x^2+3$. Note $x^2+3 = 3(\frac{x^2}{3}+1)$. So we have 
$$ \int \frac{1}{x^2+3}dx = \int \frac{1}{3(\frac{x^2}{3}+1)}dx = \frac{1}{3} \int \frac{1}{\frac{x^2}{3}+1}dx  $$
Now we want to make use of substitution rule. In doing so, we want $u^2 = \frac{x^2}{3}$, which means we can let $u = \frac{x}{\sqrt 3}$. After a bit of algebra you can conclude
$$ \int \frac{1}{x^2+3}dx = \frac{\sqrt 3}{3} \arctan( \frac{\sqrt 3 \ x}{3}) + C $$ 
