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The value of the integral $\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$ is : |
$\frac{e^x}{x}+C$, where C is a constant $\frac{e^{-x}}{x}+C$, where C is a constant $e^x+C$, where C is a constant $e^{-x}+C$, where C is a constant |
$\frac{e^x}{x}+C$, where C is a constant |
$I = \int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$ this of the type $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$ here $f(x) = \frac{1}{x}~~~~~~f'(x) = \frac{1}{x^2}$ $I = \frac{e^x}{x}+C$ |
