Integration By Parts

**My Little Pony** · February 18th 2009, 09:42 PM

Use integration by parts to find the integral:

**kalagota** · February 18th 2009, 10:14 PM

Originally Posted by My Little Pony

Use integration by parts to find the integral:

for the first time, let $u = (\ln (4x))^2$ and $dv=dx$

and for the second time, let $u= \ln 4x$ and $dv=dx$

**Mr. Engineer** · February 18th 2009, 10:21 PM

Integration by parts;

$\int u \, dv=uv - \int v \, du$

Function:
$\int(ln(4x))^2$

let $u=(ln(4x))^2$
$dv=dx$

$du=2(ln(4x))\frac{1}{4x}$
$v=x$

= $x(ln(4x))^2-\int2x(ln(4x))\frac{1}{4x}$

Then simplify the remaining integral and integrate once again

**My Little Pony** · February 19th 2009, 06:25 AM

Originally Posted by Mr. Engineer

Integration by parts;

$\int u \, dv=uv - \int v \, du$

Function:
$\int(ln(4x))^2$

let $u=(ln(4x))^2$
$dv=dx$

$du=2(ln(4x))\frac{1}{4x}$
$v=x$

= $x(ln(4x))^2-\int2x(ln(4x))\frac{1}{4x}$

Then simplify the remaining integral and integrate once again

Are you sure it's 1/4x and not 1/x?

**Mr. Engineer** · February 19th 2009, 10:20 AM

my bad it is 1/x

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