Evaluate the integral: int frac{(x+1)(x+log x | vedclass

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Question

(A) Let $I = \int \frac{(x+1)(x+\log x)^2}{x} dx$. Rewrite the integrand as: $I = \int \left( \frac{x}{x} + \frac{1}{x} \right) (x + \log x)^2 dx = \i

Vedclass · Accepted Answer

$\frac{(x+\log x)^3}{3} + c$

Vedclass · Answer

(A) Let $I = \int \frac{(x+1)(x+\log x)^2}{x} dx$. Rewrite the integrand as: $I = \int \left( \frac{x}{x} + \frac{1}{x} \right) (x + \log x)^2 dx = \int (1 + \frac{1}{x}) (x + \log x)^2 dx$. Let $u = x + \log x$. Then, $du = (1 + \frac{1}{x}) dx$. Substituting these into the integral: $I = \int u^2 du$. Integrating with respect to $u$: $I = \frac{u^3}{3} + c$. Substituting back $u = x + \log x$: $I = \frac{(x + \log x)^3}{3} + c$.

Evaluate the integral: $\int \frac{(x+1)(x+\log x)^2}{x} dx$

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