<s>[INST] Evaluate $\int \frac{1}{3+4\tan x} \, dx$ I'm am trying to evaluate $$\int \frac{1}{3+4\tan x} \, dx.$$
I tried to use the universal trig substitution, that is 
$$t = \tan{x \over 2}, \quad \tan x = {2t \over 1-t^2}, \quad dx = {2 \over t^2 + 1}dt$$
and after manipulation I got the integral to be 
$$\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt.$$
Is there a more clever approach to this problem or do I have to continue and face partial fractions? 
 [/INST] For sure, as shown in comments and answers, the tangent half-angle substitution is not the easiest way for this problem. However, computing $$I=\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt$$ is doable noticing that the roots of the denominator are $-\frac 13$, $i$,$-i$ and $3$. Using it, by partial fraction decomposition, we can arrive to $$ \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3}=-\frac{2 (4 t-3)}{25 \left(t^2+1\right)}+\frac{4}{25 (t-3)}+\frac{12}{25 (3 t+1)}$$ and get $$I=\frac{6}{25} \tan ^{-1}(t)-\frac{4}{25} \log \left(t^2+1\right)+\frac{4}{25} \log (t-3)+\frac{4}{25} \log (75 t+25)$$ and simplifying $$I=\frac{2}{25} \left(3 \tan
   ^{-1}(t)+2\log \left(\frac{25 (t-3) (3 t+1)}{t^2+1}\right)\right)$$
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