Practicing Success
Practicing Success
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The value of the integral $∫e^x(\log\, x+\frac{1}{x})dx$ is: |
$e^xlog\, x+C$, Where C is a constant. $e^{-x}log\, x+C$, Where C is a constant. $\frac{e^x}{x}+C,$ Where C is a constant. $\frac{e^{-x}}{x}+C,$ Where C is a constant. |
$e^xlog\, x+C$, Where C is a constant. |
The correct answer is Option (1) → $e^x\log\, x+C$, Where C is a constant. $∫\left(log\, x+\frac{1}{x}\right)^{e^x}dx$ $=e^x\log x+C$ as $f(x)=\log x$ $f'(x)=\frac{1}{x}$ so $\int e^x(f(x)+f'(x))dx$ $=e^xf(x)+C$ |
