How do you integrate $int ln 4x dx$ using integration by parts?

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Answer 1 · 2018-04-06T08:44:24

$=xln4x-x+C$

$intln4xdx$

You can't integrate $ln4x$ so you pick

$u=ln4x$ $rarr$ $du=1/xdx$
$dv=dx$ $rarr$ $v=x$

And by using the formula:
$intudv=uv-intvdu$

$intln4xdx=xln4x-intx/xdx$

$=xln4x-int1dx$

$=xln4x-x+C$

Answer 2 · 2018-04-06T08:44:43

The answer is $=x(ln(|4x|)-1)+C$

$intuv'=uv-intu'v$

Here,

$u=ln4x$ , $=>$ , $u'=1/(4x)*4=1/x$

$v'=1$ , $=>$ , $v=x$

Therefore,

$intln4x=xln4x-int1/x*xdx$

$=xln4x-x+C$

$=x(ln(|4x|)-1)+C$

How do you integrate int ln 4x dx int ln 4x dx using integration by parts?