Question

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

Answer 1

`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`

Answer 2

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`