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3x23logx3+4dx=____

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a
32logx3+42+C
b
3logx3+4+C
c
x3+4
d
x3+41+log31+log3+C

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detailed solution

Correct option is D

I=∫3x23log⁡x3+4dx Let x3+4=t⇒3x2dx=dtI=∫3log⁡tdt=∫tlog⁡3dt∵alog⁡xb=blog⁡xa=tlog⁡3+11+log⁡3+C=x3+41+log⁡31+log⁡3+C


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