Question

$\int \left[sin\left(logx\right)+cos\left(logx\right)\right]dx$ is equal to

• A. $xcos(logx)+c$
• B. $cos(logx)+c$
• C. $xsin(logx)+c$
• D. $sin(logx)+c$

Answer 1

Answer: (C)

$\int [\sin (\log x)+\cos (\log x)]dx$
$=\int \frac{d}{dx}\{x\sin (\log x)\}dx$
$=x\sin (\log x)+c$