Question:
Evaluate: $\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x$
Solution:
Given,
$\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x$
Let $3 x^{2}+\sin 6 x=t$
$\Rightarrow \frac{d}{d x}\left(3 x^{2}+\sin 6 x\right)=d t$
$\Rightarrow 6 x+\cos 6 x \cdot 6=d t$
$\Rightarrow x+\cos 6 x=\frac{d t}{6}$
Substituting the values,
$=\int \frac{1}{6 t} d t$
$=\frac{1}{6} \log t+c$
$=\frac{1}{6} \log \left(3 x^{2}+\sin 6 x\right)+c$