If `int (3sinx+2cosx)/ (3cosx+2sinx) dx=a x+b ln|2sinx+3cosx|+C`, then (a)`a=-(12)/(13),b=(15)/(39)` (b) `a=-7/(13),b=6/(13)` (c)`a=(12)/(13),b=(15)/(39)` (d) `a=-7/(13),b=-6/(13)`

Explanation

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`