Consider the given integral.
I=∫3x√1−x2dx
Let t=1−x2
dtdx=0−2x
−dt2=xdx
Therefore,
I=−32∫1√tdt
I=−32(2√t)+C
I=−3√t+C
On putting the value of t, we get
I=−3√1−x2+C
Hence, this is the answer.