How do you integrate #int ln(3x)# by parts?
2 Answers
Mar 12, 2017
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
Explanation:
Let
Therefore,