Find $int ln^2x dx$ ?

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Answer 1 · 2017-11-28T18:53:54

$int \ ln^2x \ dx = x^2lnx-2xlnx+2x + C$

We seek:

$I = int \ ln^2x \ dx$

Let ${ (u,=ln^2x, => (du)/dx,=(2lnx)/x), ((dv)/dx,=1, => v,=x ) :}$

Then plugging into the IBP formula:

$int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx$

We have:

$int \ (ln^2x)(1) \ dx = (ln^2x)(x) - int \ (x)((2lnx)/x) \ dx$
$:. I = x^2lnx-2int \ lnx \ dx$

Now consider the last integral:

$I_1 = int \ x^2sin2x \ dx$

We can again apply integration by parts a second time for the second integral:

Let ${ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1, => v,=x ) :}$

Then plugging into the IBP formula:

$int \ (lnx)(1) \ dx = (lnx)(x) - int \ (x)(1/x) \ dx$
$:. int \ lnx \ dx = xlnx - int \ dx$
$:. int \ lnx \ dx = xlnx - x$

Using this result along with the earlier result we find:

$I = x^2lnx-2(xlnx-x) + C$
$\ \ = x^2lnx-2xlnx+2x + C$

Find int ln^2x dx int ln^2x dx ?