How do you integrate #int ln(3x)# by parts?

2 Answers
Mar 12, 2017

#int ln(3x) dx = x(ln(3x) - 1) + C#

Explanation:

Integration by parts tells us that:

#int u(x)v'(x) dx = u(x)v(x) - int v(x) u'(x) dx#

In our example, put:

#{ (u(x) = ln(3x)), (v(x) = x) :}#

Then:

#{ (u'(x) = 3*1/(3x) = 1/x), (v'(x) = 1) :}#

So we find:

#int ln(3x) dx = int u(x)v'(x) dx#

#color(white)(int ln(3x) dx) = u(x)v(x) - int v(x)u'(x) dx#

#color(white)(int ln(3x) dx) = xln(3x) - int x*1/x dx#

#color(white)(int ln(3x) dx) = xln(3x) - int 1 dx#

#color(white)(int ln(3x) dx) = xln(3x) - x + C#

#color(white)(int ln(3x) dx) = x(ln(3x) - 1) + C#

Mar 12, 2017

The answer is #=x(ln(|3x|)-1)+C#

Explanation:

Integration by parts is

#intu'vdx=uv-intuv'dx#

Let #v=ln(3x)#, #=>#, #v'=1/(3x)*3=1/x#

#u'=1#, #=>#, #u=x#

Therefore,

#intln(3x)dx=xln(3x)-int1/x*xdx#

#=xln(3x)-x+C#

#=x(ln(3x)-1)+C#