int frac{log x}{x^3} , dx = | vedclass

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(B) To solve the integral $\int \frac{\log x}{x^3} \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \log

Vedclass · Accepted Answer

$-\frac{1}{4x^2}(2\log x + 1) + c$

Vedclass · Answer

(B) To solve the integral $\int \frac{\log x}{x^3} \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \log x$ and $dv = x^{-3} \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$. Applying the formula: $\int \frac{\log x}{x^3} \, dx = (\log x)\left(-\frac{1}{2x^2}\right) - \int \left(-\frac{1}{2x^2}\right) \left(\frac{1}{x}\right) \, dx$ $= -\frac{\log x}{2x^2} + \frac{1}{2} \int x^{-3} \, dx$ $= -\frac{\log x}{2x^2} + \frac{1}{2} \left(\frac{x^{-2}}{-2}\right) + c$ $= -\frac{\log x}{2x^2} - \frac{1}{4x^2} + c$ $= -\frac{1}{4x^2}(2\log x + 1) + c$.

$\int \frac{\log x}{x^3} \, dx = $

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