MCQ
$\int {\frac{{x\,\,dx}}{{{x^2} + 4x + 5}} = } $
- A$\frac{1}{2}\log ({x^2} + 4x + 5) + 2{\tan ^{ - 1}}(x) + c$
- B$\frac{1}{2}\log ({x^2} + 4x + 5) - {\tan ^{ - 1}}(x + 2) + c$
- C$\frac{1}{2}\log ({x^2} + 4x + 5) + {\tan ^{ - 1}}(x + 2) + c$
- ✓$\frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$