CUET Preparation Today
Find an anti-derivative of the function $\frac{1}{x}, x \neq 0$ using the method of inspection. |
$x^{-2}$ $e^x$ $\log |x|$ $\frac{x^2}{2}$$ |
$\log |x|$ |
The correct answer is Option (3) → $\log |x|$ We know that $\frac{d}{dx} (\log x) = \frac{1}{x}, x > 0 \text{ and } \frac{d}{dx} [\log (-x)] = \frac{1}{-x} (-1) = \frac{1}{x}, x < 0$ Combining above, we get $\frac{d}{dx} (\log |x|) = \frac{1}{x}, x \neq 0$ Therefore, $\int \frac{1}{x} \, dx = \log |x|$ is one of the anti derivatives of $\frac{1}{x}$. |
