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Let's say that $$f(t) = \frac{\ln t}{t^3}$$ Let $$ g = \ln t$$ and $$h = t^{-3}$$ Then by product rule:

$$(hg)' = h'g +g'h$$ we have $$f'(t) = \frac{-3}{t^4} \cdot \ln t + \frac 1 t \cdot \frac1 {t^3} = \frac{1-3\ln t}{t^4}$$

now let's take the integral of that by parts we have $$\int u\, dv = uv - \int v \, du$$

Letting $$u = 1-3\ln t$$ and $$dv = \frac{1}{t^4}\,dt$$ giving us $$du = \frac{-3} t \,dt$$ and $$v = \frac{-4}{t^5}$$ we then have by the chain rule:

$$\int \frac{1-3\ln t}{t^4} \,dt = \frac{12\ln t-4}{t^5} - 12\int \frac 1 {t^6} \,dt = \frac{12\ln t-4}{t^5} +\frac{12}{5t^5}$$ This is not equal to the starting function. I understand that this is clearly indicative of an arithmetic error somewhere, but for the life of me I can not see it. Any help would be appreciated!

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    You differentiated $dv$ to get $v$, you should have integrated. – 2017-01-10

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The antiderivative of $\dfrac1{t^4} $ is $\dfrac1{-3t^3} $ and not what you had there.

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    aaand I feel silly now... that has literally burned my evening -_- – 2017-01-10
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    @AbrahamP no need for that. Everyone has those moments ;) – 2017-01-10
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    Consider accepting the answer if you feel like your issue was solved, so as to close this matter. – 2017-01-10
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    I will as soon as the system lets me... – 2017-01-10