How do you integrate #int( 3x ln(2x) dx}#?

Question

6614 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T06:35:00

#I=(3x^2)/4(ln(4x^2)-1)+C#

Here,

#I=int(3x)ln(2x)dx#

#"Using "color(blue)"Integration by Parts"#

#color(red)(int(u*v)dx=uintvdx-int((du)/(dx)intvdx)dx#

Let, #u=ln(2x)and v=3x#,we get

#(du)/(dx)=1/(2x)*2=1/x and intvdx=(3x^2)/2#

#=>I=ln(2x)(3x^2)/2-int(1/x(3x^2)/2)dx#

#=(3x^2)/2ln(2x)-3/2intxdx+c#

#=(3x^2)/2ln(2x)-3/2*x^2/2+C#

#=(3x^2)/4(2ln(2x)-1)+C#

#=(3x^2)/4(ln(2x)^2-1)+C#

#=(3x^2)/4(ln(4x^2)-1)+C#