What is $\int \frac{{ln x}^{9}}{x^{4}} d x$ $int(lnx^9)/x^4 dx$ ?

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What is $\int \frac{{ln x}^{9}}{x^{4}} d x$ $int(lnx^9)/x^4 dx$ ?

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Answer 1 · 2015-11-08T20:06:14

It is $9 [- \frac{1}{3} x^{- 3} ln x - \frac{1}{9} x^{- 3} + C]$ $9[-1/3x^-3 lnx -1/9 x^-3 +C]$ Which we may prefer to write:
$- \frac{3 ln x + 1}{x^{3}} + C$ $-(3lnx+1)/x^3+C$

Explanation:

${ln x}^{9} = 9 ln x$ $lnx^9 = 9lnx$ , so

$\int \frac{{ln x}^{9}}{x^{2}} d x = 9 \int x^{- 4} ln x d x$ $int lnx^9/x^2 dx = 9int x^(-4) lnx dx$ Now use integration by parts.

Let $u = ln x$ $u=lnx$ and $d v = x^{- 4} d x$ $dv=x^-4 dx$ ,

so $d u = x^{- 1} d x$ $du=x^(-1) dx$ and $v = - \frac{1}{3} x^{- 3}$ $v = -1/3x^-3$

And the integral becomes:

$9 [- \frac{1}{3} x^{- 3} ln x + \frac{1}{3} \int x^{- 3} x^{- 1} d x]$ $9[-1/3x^-3 lnx +1/3 int x^-3 x^-1 dx]$

$= 9 [- \frac{1}{3} x^{- 3} ln x + \frac{1}{3} \int x^{- 4} d x]$ $= 9[-1/3x^-3 lnx +1/3 int x^-4 dx]$

$= 9 [- \frac{1}{3} x^{- 3} ln x - \frac{1}{9} x^{- 3} + C]$ $= 9[-1/3x^-3 lnx -1/9 x^-3 +C]$

Which we may prefer to write:

$= - \frac{3 ln x + 1}{x^{3}} + C$ $= -(3lnx+1)/x^3+C$

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What is $\int \frac{{ln x}^{9}}{x^{4}} d x$ $int(lnx^9)/x^4 dx$ ?

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