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

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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

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

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

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

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

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

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

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

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