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

2 Answers
George C.
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

Narad T.
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

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