Evaluate the integral int frac{log x - (log x | vedclass

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Question

(B) Let $I = \int \frac{\log x - (\log x)^2 + x^2}{x^3} dx$. We can split the integral as: $I = \int \frac{\log x - (\log x)^2}{x^3} dx + \int \frac{x

Vedclass · Accepted Answer

$\frac{(\log x)^2 + 2x^2 \log x}{2x^2} + C$

Vedclass · Answer

(B) Let $I = \int \frac{\log x - (\log x)^2 + x^2}{x^3} dx$. We can split the integral as: $I = \int \frac{\log x - (\log x)^2}{x^3} dx + \int \frac{x^2}{x^3} dx$. $I = \int \frac{\log x(1 - \log x)}{x^3} dx + \int \frac{1}{x} dx$. Let $u = \frac{\log x}{x}$. Then $du = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} dx = \frac{1 - \log x}{x^2} dx$. Substituting this into the first part: $I = \int u \cdot \frac{1}{x} dx$ is not quite right,let's re-evaluate. Actually,$\int \frac{\log x(1 - \log x)}{x^3} dx = \int \frac{\log x}{x} \cdot \frac{1 - \log x}{x^2} dx$. Let $t = \frac{\log x}{x}$,then $dt = \frac{1 - \log x}{x^2} dx$. So,$\int t dt = \frac{t^2}{2} = \frac{(\log x)^2}{2x^2}$. Adding the second part $\int \frac{1}{x} dx = \log x$. Thus,$I = \frac{(\log x)^2}{2x^2} + \log x + C = \frac{(\log x)^2 + 2x^2 \log x}{2x^2} + C$.

Evaluate the integral $\int \frac{\log x - (\log x)^2 + x^2}{x^3} dx$ (where $C$ is the constant of integration).

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