Evaluate x+ 6x 3x^2+ 6x dx - Vidyadip

Question

Evaluate $\int\frac{\text{x}+\cos6\text{x}}{3\text{x}^2+\sin6\text{x}}\text{ dx}$

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Answer

Let $3\text{x}^2+\sin6\text{x}=\text{t}$
$6\text{x}+6\cos6\text{x dx}=\text{dt}$
$(\text{x}+\cos6\text{x})\text{dx}=\frac{\text{dt}}{6}$
Thus, $\text{I}=\int\frac{\text{x}+\cos6\text{x}}{3\text{x}^2+\sin6\text{x}}\text{ dx}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{6}\log|\text{t}|+\text{C}$
$=\frac{\log\big|3\text{x}^2+\sin6\text{x}\big|}{6}+\text{C}$

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