Evaluate the integral: int frac{ln(6x^2)}{x} | vedclass

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Question

(B) Let $I = \int \frac{\ln(6x^2)}{x} \, dx$. Substitute $t = \ln(6x^2)$. Then,$dt = \frac{1}{6x^2} \cdot \frac{d}{dx}(6x^2) \, dx = \frac{1}{6x^2} \c

Vedclass · Accepted Answer

$\frac{1}{4}[\ln(6x^2)]^2 + C$

Vedclass · Answer

(B) Let $I = \int \frac{\ln(6x^2)}{x} \, dx$. Substitute $t = \ln(6x^2)$. Then,$dt = \frac{1}{6x^2} \cdot \frac{d}{dx}(6x^2) \, dx = \frac{1}{6x^2} \cdot 12x \, dx = \frac{2}{x} \, dx$. This implies $\frac{dx}{x} = \frac{dt}{2}$. Substituting these into the integral: $I = \int t \cdot \frac{dt}{2} = \frac{1}{2} \int t \, dt$. Integrating with respect to $t$: $I = \frac{1}{2} \cdot \frac{t^2}{2} + C = \frac{t^2}{4} + C$. Substituting $t = \ln(6x^2)$ back: $I = \frac{1}{4} [\ln(6x^2)]^2 + C$.

Evaluate the integral: $\int \frac{\ln(6x^2)}{x} \, dx$

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