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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

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,$I = \frac{3^x}{x} + c$.

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

Similar Questions

If $\int {{e^{\sec x}}\left( {\sec x + \tan x f(x) + (\sec x \tan x + \sec^2 x)} \right)dx = {e^{\sec x}}f(x) + C}$,then a possible choice of $f(x)$ is

$\int \left( \frac{\log x - 1}{1 + (\log x)^2} \right)^2 dx = $

$\int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx = $

$\int e^x \frac{x^2+1}{(x+1)^2} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module