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Q.

∫3x23logx3+4dx=____

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a

32logx3+42+C

b

3logx3+4+C

c

x3+4

d

x3+41+log31+log3+C

answer is D.

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Detailed Solution

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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