How do you evaluate the integral $\int x^{- 3} ln x$ $int x^-3lnx$ ?

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Answer 1 · 2017-01-02T06:49:56

The answer is $= - \frac{2 ln (∣ x ∣) + 1}{4 x^{2}} + C$ $=-(2ln(∣x∣)+1)/(4x^2)+C$

We need $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C (n!=-1)$

$intu'v=uv-intuv'$

The integral is $=intx^(-3)lnxdx$

Let $u'=x^-3$ , $=>$ , $u=-x^-2/2$

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

Therefore,

$intx^(-3)lnxdx=-1/(2x^2)*lnx-int-dx/(2x^3)$

$=-lnx/(2x^2)+intdx/(2x^3)$

$=-lnx/(2x^2)-1/2*1/(2x^2)+C$

$=-lnx/(2x^2)-1/(4x^2)+C$

$=-(2ln(∣x∣)+1)/(4x^2)+C$

How do you evaluate the integral int x^-3lnx∫x−3lnxint x^-3lnx?