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? 
 A: Hint:
Let
\begin{equation}
I=\int\frac{1}{a+b\tan x}\ dx=\int\frac{\cos x}{a\cos x+b\sin x}\ dx
\end{equation}
and
\begin{equation}
J=\int\frac{\sin x}{a\cos x+b\sin x}\ dx
\end{equation}
then
\begin{equation}
aI+bJ=x+C_1\tag1
\end{equation}
and
\begin{equation}
bI-aJ=\log(a\cos x+b\sin x)+C_2\tag2
\end{equation}
Setting $a=3$ and $b=4$, then it's easy to determine $I$ by linear combinations $(1)$ and $(2)$.
A: Note that $$\frac{1}{3 + 4 \tan x} = \frac{\cos x}{3 \cos x + 4 \sin x} = f(x).$$
Next, observe that $$\log (3 \cos x + 4 \sin x) + C = \int \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx = \int g(x) \, dx,$$ via the obvious substitution $u = 3 \cos x + 4 \sin x$.  Now the goal is to find a linear combination of the integrands $f$ and $g$ such that for some real-valued scalar constants $A$, $B$, $$A f(x) + B g(x) = 1.$$  Well, this is simple:  we have $$A \cos x + B(4 \cos x - 3 \sin x) = 3 \cos x + 4 \sin x,$$ and collecting like terms, we obtain $$A + 4B = 3, \quad -3B = 4,$$ or $$A = \frac{25}{3}, \quad B = -\frac{4}{3}.$$  Therefore, $$x = \int 1 \, dx = \int A f(x) + B g(x) \, dx = A \int f(x) \, dx + B \int g(x) \, dx,$$ or $$\int f(x) \, dx = \frac{3}{25} \left( x + \frac{4}{3} \log (3 \cos x + 4 \sin x) \right) + C,$$ and the rest is trivial.
A: It is probably easier to use the substitution $x=\arctan(t)$ then find a partial fraction decomposition for
$$ g(t) = \frac{1}{(3+4t)(1+t^2)}. $$
By computing the residue at the simple poles $t\in\left\{-\frac{3}{4},i,-i\right\}$ it follows that:
$$ g(t) = \frac{4}{25}\cdot\frac{1}{x+\frac{3}{4}}+\frac{1}{25}\cdot\frac{3-4t}{1+t^2} $$
and:

$$ \int g(t)\,dt = C + \frac{1}{25}\left[4\log\left(t+\frac{3}{4}\right)+3\arctan(t)-2\log(1+t^2)\right].$$

A: 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 $$I=\frac{2}{25} \left(3 \tan
   ^{-1}(t)+2\log \left(\frac{25 (t-3) (3 t+1)}{t^2+1}\right)\right)$$
