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
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