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

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Answer 1 · 2018-04-06T08:44:24

$= x ln 4 x - x + C$ $=xln4x-x+C$

$\int ln 4 x d x$ $intln4xdx$

You can't integrate $ln 4 x$ $ln4x$ so you pick

$u = ln 4 x$ $u=ln4x$ $\to$ $rarr$ $d u = \frac{1}{x} d x$ $du=1/xdx$
$d v = d x$ $dv=dx$ $\to$ $rarr$ $v = x$ $v=x$

And by using the formula:
$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

$\int ln 4 x d x = x ln 4 x - \int \frac{x}{x} d x$ $intln4xdx=xln4x-intx/xdx$

$= x ln 4 x - \int 1 d x$ $=xln4x-int1dx$

$= x ln 4 x - x + C$ $=xln4x-x+C$

Answer 2 · 2018-04-06T08:44:43

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

$intuv'=uv-intu'v$

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 ∫ln4xdxint ln 4x dx using integration by parts?