∫3x23logx3+4dx=____
32logx3+42+C
3logx3+4+C
x3+4
x3+41+log31+log3+C
I=∫3x23logx3+4dx Let x3+4=t⇒3x2dx=dtI=∫3logtdt=∫tlog3dt∵alogxb=blogxa=tlog3+11+log3+C=x3+41+log31+log3+C