We have `I=int(3sinx+2cosx)/(3cosx+2sinx)dx` Let `3sinx+2cosx=mu(d)/(dx)(3cosx+2sinx)+lambda(3cosx+2sinx)` `=mu(-3sinx+2cosx)+lambda(3cosx+2sinx)` Comparing the coefficients of `sin x` and `cosx` on both sides, we get `-3mu+2lambda=3 " and " 2mu+3lambda=2` ` " or " lambda=(12)/(13) " and " mu=-(5)/(13) ` `:. I=int(mu(-3sinx+2cosx)+lambda(3cosx+2sinx))/(3cosx+2sinx)dx` `=lambda int 1dx+mu int(-3sinx+2cosx)/(3cosx+2sinx)dx` `=lambda x +mu int(dt)/(t), " where " t=3cosx+2sinx` `=lambda x +mu log|t|+C` `=(12)/(13)x+(-5)/(13)log|3cosx+2sinx|+C`
