CUET Preparation Today
Find $\int \frac{1}{x(1+x^2)} dx$. |
$\ln|x| + \frac{1}{2} \ln(1+x^2) + C$ $\tan^{-1}x - \ln|x| + C$ $\ln|x| - \frac{1}{2} \ln(1+x^2) + C$ $\frac{1}{2} \ln \left| \frac{x^2}{1+x^2} \right| + C$ |
$\ln|x| - \frac{1}{2} \ln(1+x^2) + C$ |
The correct answer is Option (3) → $\ln|x| - \frac{1}{2} \ln(1+x^2) + C$ $I = \int \frac{1}{x(1+x^2)} dx = \int \left( \frac{1}{x} - \frac{x}{1+x^2} \right) dx$ $= \log |x| - \frac{1}{2} \ln(1+x^2) + C$ |
