$\int \frac{2 \sin x-3 \cos x}{4 \cos x-3 \sin x} d x=$

• A

  $\frac{1}{25}[17 \log |4 \cos x-3 \sin x|-6 x]+c$
• B

  $\frac{1}{25}[x-18 \log |4 \cos x-3 \sin x|]+c$
• Correct answer
  C

  $\frac{1}{25}[\log |4 \cos x-3 \sin x|-18 x]+c$
• D

  $\frac{1}{25}[17 x-6 \log |4 \cos x-3 \sin x|]+c$

Hence, the correct answer is option: 'C'.