Evaluate the following integrals:


Evaluate: $\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x$



$\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x$

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,

$=\int \frac{1}{6 t} d t$

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

$=\frac{1}{6} \log \left(3 x^{2}+\sin 6 x\right)+c$

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