MathematicsMedium210×since 2002Q3465Let 5f(x)+4f(1x)=1x+3,x>05 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 05f(x)+4f(x1)=x1+3,x>0. Then 18 \int_\limits{1}^{2} f(x) d x is equal to :A10loge2+610 \log _{\mathrm{e}} 2+610loge2+6B5loge2−35 \log _{e} 2-35loge2−3C10loge2−610 \log _{\mathrm{e}} 2-610loge2−6D5loge2+35 \log _{\mathrm{e}} 2+35loge2+3Check answerSkip