Mathematics

# Solve: $\displaystyle \int \dfrac{\log^x}{x^3}dx$

##### SOLUTION
$\displaystyle\int \dfrac{\log x}{x^3}dx$
$=\displaystyle\int e^{-2t}tdt$ Let $\log x=t$ $\Rightarrow x=e^{t} \dfrac{dx}{x}=dt$
$=t\cdot\displaystyle\int e^{-2t}dt-\displaystyle\int \left(\dfrac{d(t)}{dt}\displaystyle\int e^{-2tdt}\right)dt$
$=\dfrac{te^{-2t}}{(-2)}-\displaystyle\int \dfrac{1e^{-2t}}{(-2)}dt$
$=\dfrac{-te^{-2t}}{2}-\dfrac{e^{-2t}}{4}+c$
$=\dfrac{e^{-2t}}{2}\left(-t-\dfrac{1}{2}\right)+c$
$=\dfrac{e^{-2\log x}}{2}\left(-\log x-\dfrac{1}{2}\right)+c$
$=\dfrac{1}{2x^2}\left(-\log x-\dfrac{1}{2}\right)+c$ $\left(e^{\log x-2}=x^{-2}\right)$
$\therefore \displaystyle\int \dfrac{\log x}{x^3}dx=\dfrac{1}{2x^2}\left(-\log x-\dfrac{1}{2}\right)+c$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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