Question

evaluate 
integral of x+cos 6 x/3 x²+sin 6 x
please ans with detail

Answer 1

$Integral\ of\ \frac{x+Cos 6x}{3x^2+Sin 6x}=\frac{u}{v}\\ \\Looking\ at\ denominator, \frac{dv}{dx}=6x+6 Cos6x\\ \\u= \frac{1}{6} \frac{dv}{dx}\\ \\ \int\limits^{}_{} {\frac{x+Cos 6x}{3x^2+Sin 6x}} \, dx =\frac{1}{6} \int\limits^{}_{} {\frac{1}{v}} \, dv =\frac{1}{6}Ln\ v=\frac{1}{6}Ln(3x^2+Sin6x)$