How do you integrate $\int \frac{1}{x {(ln x)}^{3}}$ $int 1/(x(lnx)^3)$ by parts?

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How do you integrate $\int \frac{1}{x {(ln x)}^{3}}$ $int 1/(x(lnx)^3)$ by parts?

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Answer 1 · 2016-12-01T23:15:44

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Answer 2 · 2016-12-01T23:24:01

If I had a choice, I would not use parts, I would use the substitution $u = {(ln x)}^{- 3}$ $u = (lnx)^-3$

Explanation:

If I was told I needed to use parts, I would take $u = 1$ $u = 1$ so $d u = 0 d x$ $du = 0dx$ and $d v = {(ln x)}^{- 3} \frac{1}{x} d x$ $dv = (lnx)^-3 1/x dx$ so $v = - \frac{1}{2} {(ln x)}^{- 2}$ $v = -1/2(lnx)^-2$ (by sustitution).

$u v - \int v d u = - \frac{1}{2 {(ln x)}^{2}} - \int 0 d x$ $uv-int v du = -1/(2(lnx)^2) - int 0 dx$

$= - \frac{1}{2 {(ln x)}^{2}} + C$ $= -1/(2(lnx)^2) +C$

(I know it isn't " really" integration by parts.)

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How do you integrate $\int \frac{1}{x {(ln x)}^{3}}$ $int 1/(x(lnx)^3)$ by parts?

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