Solve the following integration:
\[\int\frac{x+cosx}{3x^{2}+6sinx}dx\]

\(\int\frac{x+cosx}{3x^{2}+6sinx}\)\(dx\)

Let, \(3x^{2}+6sinx\) \(= t\)

Differentiating w.r.t \(x\)

\(6x+6cosx\) \(dx\) \(=dt\)

\(6(x+cosx)\) \(dx\) \(=dt\)

\((x+cosx)\)\(dx\)\(=\)\(\int\frac{1}{6}\)\(dt\)

\(\int\frac{x+cosx}{3x^{2}+6sinx}\)\(=\)\(\int\frac{1}{6}\)\(\frac{dt}{t}\)

\(=\)\(\frac{1}{6}\)\(\int\frac{dt}{t}\)

\(=\)\(\frac{1}{6}\)\(\log|t|+C\)

Putting the value of \(t\) = \(3x^{2}+6sinx\)

\(=\)\(\frac{1}{6}\)\(\log|3x^{2}+6sinx|+C\)