How do you integrate $\int x {(ln x)}^{2}$ $int x(lnx)^2$ using integration by parts?

Question

1718 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-20T14:27:30

The answer is $= \frac{x^{2}}{2} {(ln x)}^{2} - \frac{x^{2}}{2} ln x + \frac{x^{2}}{4} + C$ $=x^2/2(lnx)^2-x^2/2lnx+x^2/4+C$

$intuv'=uv-intu'v$

Here,

$u=(lnx)^2$ , $=>$ , $du=2/xlnx$

$v'=x$ , $=>$ , $v=x^2/2$

Therefore, the integral is

$intx(lnx)^2dx=x^2/2(lnx)^2-int2/xlnx*x^2/2dx$

$=x^2/2(lnx)^2-intxlnxdx$

To calculate $intxlnxdx$ , perform the integration by parts

$u=lnx$ , $=>$ , $u'=1/x$

$v'=x$ , $=>$ , $v=x^2/2$

Therefore,

$intxlnxdx=x^2/2lnx-int1/x*x^2/2dx$

$=x^2/2lnx-1/2intxdx$

$=x^2/2lnx-x^2/4$

Putting all together

$intx(lnx)^2dx=x^2/2(lnx)^2-x^2/2lnx+x^2/4+C$

How do you integrate int x(lnx)^2 ∫x(lnx)2int x(lnx)^2 using integration by parts?