$\int \frac{1}{(x+2)(1+x)^2} \mathrm{~d} x$ has the value
- $2 \log \left(\frac{x+2}{x^2+1}\right)+4 \tan ^{-1} x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\log \frac{x+2}{x^2+1}-4 \tan ^{-1} x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\log \frac{(x+2)^2}{\left(x^2+1\right)}+4 \tan ^{-1} x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\log \frac{(x+2)}{\left(x^2+1\right)^2}-4 \tan ^{-1} x+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution
Asked in: MHT CET 2023 (12 May Shift 2)