[tex]\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}[/tex]
A standard approach would be the tangent half-angle substitution:
[tex]t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx[/tex]
Then
[tex]\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}[/tex]
[tex]\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}[/tex]
from which we get
[tex]\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt[/tex]
So the integral becomes
[tex]\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}[/tex]
Rewrite the denominator as
[tex]2t^2-3t-2=(2t+1)(t-2)[/tex]
and expand the integrand into its partial fractions:
[tex]\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)[/tex]
We have
[tex]\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C[/tex]
[tex]=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C[/tex]
[tex]=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C[/tex]
