We can do this in a number of ways.
First, let's get rid of that three using a log property:
ln(3x) = ln(3) + ln(x) ln(3x)=ln(3)+ln(x)
ln(3) is a constant, so its antiderivative will just be xln(3)xln(3) which is fine to deal with later.
For ln(x)ln(x), we can write the following:
dy/dx = ln(x) rightarrow dy = ln(x) dxdydx=ln(x)→dy=ln(x)dx
Let's use u substitution with u=ln(x)u=ln(x), i.e. du = 1/x dx = e^(-u) dxdu=1xdx=e−udx. This yields
dy = u e^u du implies y = int u e^u du dy=ueudu⇒y=∫ueudu
By integration by parts,
y = ue^u - int e^u du = ue^u - e^u + C y=ueu−∫eudu=ueu−eu+C
Putting back xx,
y = xlnx - x + C y=xlnx−x+C
This yields the final antiderivative as
int\ ln(3x)dx = xln(3) + xln(x) - x + C = xln(3x) - x + C
