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

Question

William Abdalla · Accepted Answer

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

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#

Isaiah Allen · Answer

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

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#

William Abdalla · Answer

undefined To integrate $ \int \ln(3x) \, dx $ by parts, you can use the integration by parts formula, which states:

$$ \int u \, dv = uv - \int v \, du $$

Let $ u = \ln(3x) $ and $ dv = dx $.

Then, $ du = \frac{1}{x} \, dx $ and $ v = x $.

Now, applying the integration by parts formula:

$$ \int \ln(3x) \, dx = x\ln(3x) - \int x \cdot \frac{1}{x} \, dx $$

$$ = x\ln(3x) - \int dx $$

$$ = x\ln(3x) - x + C $$

Therefore, the integral of $ \ln(3x) $ with respect to $ x $ is $ x\ln(3x) - x + C $, where $ C $ is the constant of integration.