int frac{3^x(x log 3-1)}{x^2} d x= | vedclass

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(A) Let $I = \int \frac{3^x(x \log 3 - 1)}{x^2} dx$. We can rewrite the integrand as: $I = \int \left( \frac{x \cdot 3^x \log 3}{x^2} - \frac{3^x}{x^2

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$\frac{3^x}{x}+c$

Vedclass · Answer

(A) Let $I = \int \frac{3^x(x \log 3 - 1)}{x^2} dx$. We can rewrite the integrand as: $I = \int \left( \frac{x \cdot 3^x \log 3}{x^2} - \frac{3^x}{x^2} \right) dx$ $I = \int \left( \frac{3^x \log 3}{x} - \frac{3^x}{x^2} \right) dx$. Recall the derivative rule for a quotient or product. Let $f(x) = \frac{3^x}{x}$. Then $f'(x) = \frac{x \cdot \frac{d}{dx}(3^x) - 3^x \cdot \frac{d}{dx}(x)}{x^2}$. Since $\frac{d}{dx}(3^x) = 3^x \log 3$, we have: $f'(x) = \frac{x \cdot 3^x \log 3 - 3^x \cdot 1}{x^2} = \frac{3^x(x \log 3 - 1)}{x^2}$. Thus, $\int f'(x) dx = f(x) + c$. Therefore, $I = \frac{3^x}{x} + c$.

$\int \frac{3^x(x \log 3-1)}{x^2} d x=$

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