Question

Integral of (x+cos6x) / (3x 2 +sin6x) dx

Answer 1

let I = $\int\limits \frac{ x + cos6x }{3 x^{2} + sin 6x} dx$

let $3 x^{2} + sin 6x = t$

$(6x+6cos 6x) dx = dt$

$(x + cos 6x) dx = dt/6$

$I = \frac{1}{6} \int\limits \frac{dt}{t}$

$= \frac{1}{6} log |t| +c$

$= \frac{1}{6} log |3 x^{2} + sin 6x| + c$