The correct option is B 17ln∣∣∣tanx−3tanx+4∣∣∣+C
Solving the intregral of the form I=∫dxacos2x+bsin2x+cosx sinx
requires multiplying both numerator and denominator of the integrand by sec2x and substituting t=tanx
Here, multiplying the numerator and denominator of the integrand by sec2x we get,
I=∫sec2x dxsin2x sec2x+cosx sinx sec2x−12cos2x sec2x ⇒I=∫sec2x dxtan2x+tanx−12
Now, substituting t=tanx
we get, dt=sec2x dx
Substituting these back in the integral we get,
I=∫dtt2+t−12⇒I=∫dt(t−3)(t+4)⇒I=17∫(1t−3−1t+4]dt⇒I=17(ln(t−3|−ln(t+4|]+C⇒I=17ln(t−3t+4∣∣+C
Now, substituting back t=tanx
we get I=17ln(tanx−3tanx+4∣∣+C
Thus, Option b. is the correct.