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

Correct Answer is : 125[log|4cosx−3sinx|−18x]+C

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

\frac{1}{25}[x-18 \log |4 \cos x-3 \sin x|]+C

\frac{1}{25}[\log |4 \cos x-3 \sin x|-18 x]+C

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

TG EAMCET 2 May 2025 Shift 1 Paper

Question : 72 of 160

Marks: +1, -0

∫‌

2sin‌x−3‌cos‌x

4‌cos‌x−3sin‌x

‌dx=

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