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The integral $∫e^x\left(1+\frac{1}{x}+log\, x\right)$ is equal to : |
$e^x(x+log \, x) + C$ $e^x(1+log \, x) + C$ $e^xlog\, x + C$ $e^x(x-log \, x) + C$ |
$e^x(1+log \, x) + C$ |
The correct answer is Option (2) → $e^x(1+log \, x) + C$ $∫e^x+e^x\left(1+\frac{1}{x}+\log x\right)$ $=e^x+∫e^x\left(1+\frac{1}{x}+\log x\right)$ $⇒e^x+e^x\log x+C$ $=e^x(1+\log x)+C$ |
